Class 6 Maths Practice- Basic Geometrical Ideas

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📐 Class 6 Maths | Basic Geometrical Ideas | Timer Quiz (60s)

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Class 6 Maths Practice- Playing with Numbers

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Class 6 Maths Practice- Whole Numbers

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Class 6 Maths Practice- Knowing Our Numbers

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10th Polynomials-Project Works

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Polynomial Projects - Student Submission

Polynomial Projects

Mathematics Project Work - Class X

Name	Rahul Sharma
Class	X - A
Roll Number	25
Lesson	Polynomials
Academic Year	2023-2024

Project 1: Exploring the Parabola - A Graphical Journey into Quadratic Polynomials

Aim

To understand the relationship between the coefficients of a quadratic polynomial (ax² + bx + c) and the shape/position of its graph (a parabola), and to visually interpret its zeroes.

Objectives

To plot graphs of various quadratic polynomials
To analyze the effect of coefficients on the parabola's shape and position
To identify zeroes of polynomials from their graphs
To verify the relationship between zeroes and coefficients

Materials

Graph paper
Pencil and eraser
Scale/Ruler
Different colored pens

Tools

Mathematical compass (optional)
Calculator

Procedure

Selected various quadratic polynomials with different coefficient values
Created value tables for each polynomial by substituting x values
Plotted points on graph paper for each polynomial using appropriate scale
Joined the points with smooth curves to form parabolas
Analyzed the direction, width, and position of each parabola
Identified x-intercepts (zeroes) from the graphs
Verified the zeroes using algebraic methods
Compared the sum and product of zeroes with coefficient relationships

Sample Graph: y = x² + 2x - 3

[Graph showing parabola with vertex at (-1,-4) and zeroes at x=-3 and x=1]

Observations

When a > 0, the parabola opens upwards; when a < 0, it opens downwards
Larger |a| values produce narrower parabolas
The constant term c determines the y-intercept of the graph
The coefficient b affects the position of the vertex along the x-axis
The number of x-intercepts (zeroes) can be 0, 1, or 2
For y = x² + 2x - 3, the zeroes are x = -3 and x = 1
Sum of zeroes = -2 = -b/a, Product of zeroes = -3 = c/a

Polynomial	Zeroes	Sum of Zeroes	Product of Zeroes
x² + 2x - 3	-3, 1	-2	-3
x² - 5x + 6	2, 3	5	6
2x² - 8x + 6	1, 3	4	3

Conclusion

The graph of a quadratic polynomial is always a parabola. The coefficient a determines the direction and width of the parabola, b affects the position of the vertex, and c gives the y-intercept. The zeroes of the polynomial are the x-coordinates where the graph intersects the x-axis. The relationships α+β = -b/a and αβ = c/a were verified through this project.

Experience

This project helped me visualize the abstract concept of polynomials. Drawing the graphs made it easier to understand how changing coefficients affects the shape and position of the parabola. I learned to connect algebraic expressions with their geometric representations. The hands-on approach made the mathematical relationships more memorable and intuitive.

Quadratic Polynomial Calculator

Coefficient a:

Coefficient b:

Coefficient c:

Project 2: The Algebra-Geometry Bridge: Zeroes and Coefficients of Cubic Polynomials

Aim

To verify the relationship between the zeroes and coefficients of a cubic polynomial and to use the Division Algorithm to find all zeroes when one is known.

Objectives

To verify relationships between zeroes and coefficients of cubic polynomials
To apply the Division Algorithm to factorize polynomials
To find all zeroes of a cubic polynomial when one zero is known
To connect algebraic and geometric representations of polynomials

Materials

Pen and paper
Calculator
Chart paper for presentation

Tools

Mathematical tables (if needed)

Procedure

Selected cubic polynomials that can be factorized easily
Found zeroes of each polynomial by factorization method
Recorded the zeroes and coefficients for each polynomial
Verified the relationships:
- α + β + γ = -b/a
- αβ + βγ + γα = c/a
- αβγ = -d/a
Selected a cubic polynomial with one known zero
Applied the Division Algorithm to divide the polynomial by (x - known zero)
Factorized the quotient to find the remaining zeroes
Verified all zeroes by substituting in the original polynomial

Cubic Polynomial Graph: y = x³ - 4x

[Graph showing cubic curve with zeroes at x=-2, x=0, and x=2]

Observations

Cubic polynomials can have 1 or 3 real zeroes
The relationships between zeroes and coefficients hold for all cubic polynomials
The Division Algorithm is an efficient method to find all zeroes when one is known
The graph of a cubic polynomial can intersect the x-axis at 1 or 3 points
For x³ - 4x, the zeroes are -2, 0, and 2
Sum of zeroes = 0 = -b/a, Sum of products = -4 = c/a, Product = 0 = -d/a

Polynomial	Zeroes (α, β, γ)	α+β+γ	αβ+βγ+γα	αβγ
x³ - 4x	-2, 0, 2	0	-4	0
x³ - 6x² + 11x - 6	1, 2, 3	6	11	6
2x³ - 5x² - 14x + 8	4, -2, ½	2.5	-7	-4

Conclusion

Cubic polynomials follow specific relationships between their zeroes and coefficients. The Division Algorithm provides a systematic method to find all zeroes when one zero is known. These relationships and methods connect the algebraic and geometric aspects of polynomials, enhancing our understanding of their behavior and properties.

Experience

Working with cubic polynomials helped me understand higher-degree algebraic expressions. The Division Algorithm was particularly interesting as it provided a structured approach to polynomial factorization. Seeing the consistent relationships between zeroes and coefficients across different examples reinforced my understanding of polynomial theory. This project strengthened my algebraic manipulation skills and deepened my appreciation for the interconnectedness of mathematical concepts.

Cubic Polynomial Calculator

Coefficient a:

Coefficient b:

Coefficient c:

Coefficient d:

Acknowledgement

I would like to express my sincere gratitude to my mathematics teacher, Mrs. Priya Mehta, for her guidance and support throughout this project. Her explanations and insights helped me understand the concepts clearly. I also thank my school for providing the necessary resources and my parents for their encouragement. Finally, I acknowledge the textbook authors whose work formed the foundation of this project.

References

Mathematics Textbook for Class X - NCERT
Comprehensive Mathematics - R.D. Sharma
Mathematics Class X - R.S. Aggarwal
Khan Academy - Online resources on polynomials
Byju's Learning App - Interactive polynomial lessons

10th Sets-Project Works

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Sets Projects - Mathsstudent.com

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Venn Diagrams and Set Operations-Project-1

Name: K. Syam

Class: X

Roll Number: 10

Lesson: Sets

Subject: Mathematics

Aim

To understand and visualize set operations using Venn diagrams and to solve practical problems using set theory concepts.

Objectives

To understand the concept of sets, subsets, and universal sets
To learn and apply set operations: union, intersection, and difference
To represent sets and set operations using Venn diagrams
To solve real-world problems using set theory

Materials & Tools

Materials Used: Chart paper, colored pens, ruler, compass, scissors

Tools: Mathematical calculation, Venn diagram representation

Procedure

Collected data from classmates about their preferences for different subjects
Defined universal set as all students in the class
Created sets for students liking Mathematics, Science, and English
Drew Venn diagrams to represent the relationships between these sets
Calculated union, intersection, and difference of the sets
Analyzed the results to find patterns and relationships

Observations & Data Analysis

Survey Data Collected:

Total students in class: 40

Students who like Mathematics: 25
Students who like Science: 22
Students who like English: 18
Students who like both Mathematics and Science: 12
Students who like both Mathematics and English: 8
Students who like both Science and English: 6
Students who like all three subjects: 4

Set Operations:

Let M = Set of students who like Mathematics

Let S = Set of students who like Science

Let E = Set of students who like English

Set Operation	Calculation	Result
M ∪ S (Union)	n(M) + n(S) - n(M∩S) = 25 + 22 - 12	35
M ∪ E (Union)	n(M) + n(E) - n(M∩E) = 25 + 18 - 8	35
S ∪ E (Union)	n(S) + n(E) - n(S∩E) = 22 + 18 - 6	34
M ∩ S (Intersection)	Given in survey	12
M ∩ E (Intersection)	Given in survey	8
S ∩ E (Intersection)	Given in survey	6
M ∩ S ∩ E (Intersection)	Given in survey	4

Conclusion

Through this project, I learned that Venn diagrams are powerful visual tools for understanding set relationships. The survey data analysis showed that many students have overlapping interests in different subjects. Set operations like union, intersection, and difference helped quantify these relationships. The principle of inclusion-exclusion was particularly useful in calculating union of sets without double-counting common elements.

Experience of the Student

This project helped me understand how abstract mathematical concepts like sets have practical applications in organizing and analyzing real-world data. Creating Venn diagrams made the relationships between sets much clearer than just working with formulas. I also learned how to conduct a survey and present the results mathematically.

Acknowledgement

1. I thank our Mathematics teacher for guiding us through the concepts of set theory

2. I appreciate my classmates for actively participating in the survey

Reference Books / Resources

X Class State Board Mathematics Textbook - Chapter 2: Sets
IX Class Mathematics Textbook
Online resources about Venn diagrams and set operations

Mathsstudent.com

Finite and Infinite Sets in Daily Life- Project -2

Name: K. Smalhor2

Class: X

Roll Number: 10

Lesson: Sets

Subject: Mathematics

Aim

To identify and classify various collections as finite or infinite sets and to understand the concept of cardinality in finite sets.

Objectives

To differentiate between finite and infinite sets
To identify empty sets and understand their properties
To calculate cardinality of finite sets
To find subsets of given finite sets
To apply set concepts to real-life collections

Materials & Tools

Materials Used: Notebook, pen, ruler, collection of various objects

Tools: Observation, classification, mathematical calculation

Procedure

Collected various groups of objects from daily life
Classified each collection as a set
Determined whether each set is finite or infinite
For finite sets, calculated their cardinality
Found all possible subsets of smaller finite sets
Identified empty sets in practical situations

Observations & Analysis

Classification of Sets:

Set Description	Roster Form	Type	Cardinality
Days of the week	{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}	Finite	7
Prime numbers less than 20	{2, 3, 5, 7, 11, 13, 17, 19}	Finite	8
Vowels in English alphabet	{a, e, i, o, u}	Finite	5
Natural numbers	{1, 2, 3, 4, ...}	Infinite	Not defined
Points on a line segment	Cannot be listed	Infinite	Not defined
Even prime numbers greater than 2	{ }	Empty	0

Subsets of a Finite Set:

Let A = {1, 2, 3}

All subsets of A:

Empty set: { } or ∅
Single element subsets: {1}, {2}, {3}
Two element subsets: {1, 2}, {1, 3}, {2, 3}
The set itself: {1, 2, 3}

Total number of subsets = 2³ = 8

Cardinality of Union and Intersection:

Let B = {2, 4, 6, 8, 10} and C = {1, 2, 3, 4, 5}

n(B) = 5, n(C) = 5
B ∩ C = {2, 4}, so n(B ∩ C) = 2
B ∪ C = {1, 2, 3, 4, 5, 6, 8, 10}, so n(B ∪ C) = 8
Verification: n(B ∪ C) = n(B) + n(C) - n(B ∩ C) = 5 + 5 - 2 = 8

Conclusion

This project helped me understand the classification of sets as finite, infinite, or empty. Finite sets have a definite number of elements that can be counted, while infinite sets continue indefinitely. The empty set is a special case with no elements. The concept of cardinality is applicable only to finite sets, and the formula for the number of subsets (2ⁿ where n is the cardinality) is a powerful tool in set theory. These concepts have practical applications in data organization, database management, and problem-solving.

Experience of the Student

Working on this project made me more observant about collections in daily life. I started seeing sets everywhere - from utensils in the kitchen to books on my shelf. Calculating all subsets of a set was challenging but interesting, and I discovered the pattern that the number of subsets doubles with each additional element. The concept of infinite sets was initially difficult to grasp, but thinking about examples like the set of all points on a line helped me understand it better.

Acknowledgement

1. I thank our Mathematics teacher for explaining the concepts of finite and infinite sets clearly

2. I appreciate my family members for helping me collect various objects for classification

Reference Books / Resources

X Class State Board Mathematics Textbook - Chapter 2: Sets
IX Class Mathematics Textbook
Online resources about finite and infinite sets

10th Real Numbers- Project Works

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Real Numbers Projects - Mathsstudent.com

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Euclid's Algorithm in Action!

Name: K. Arun

Class: X

Roll Number: 10

Lesson: Real Numbers

Subject: Mathematics

Aim

To find the Highest Common Factor (HCF) of given pairs of numbers using Euclid's Division Algorithm and to verify it through a practical activity.

Objectives

To understand and apply Euclid's Division Lemma
To find HCF using Euclid's Algorithm
To verify the algorithm with a practical activity using paper strips

Materials & Tools

Materials Used: Paper strips (two different colours), scale, pen, pencil, sketch pens, scissors

Tools: Mathematical calculation, Practical experiment

Procedure

Two pairs of numbers were selected: (50, 75) and (32, 96)
For each pair, Euclid's Division Algorithm was applied step-by-step until the remainder became zero
To verify the result for (50, 75), two paper strips of lengths 50 cm and 75 cm were cut
The longer strip (75 cm) was measured using the shorter strip (50 cm)
The leftover part (25 cm) was used to measure the 50 cm strip
The process continued until no remainder was left

Observations & Calculations

A) For numbers 50 and 70:

70 = 50 × 1 + 20
50 = 20 × 2 + 10
20 = 10 × 2 + 0
HCF (50, 70) = 10

B) For numbers 32 and 96:

96 = 32 × 3 + 0
HCF (32, 96) = 32

Application of Euclid's Algorithm

Numbers (a, b)	Division Steps (a = bq + r)	HCF
50, 70	70 = 50 × 1 + 20 50 = 20 × 2 + 10 20 = 10 × 2 + 0	10
32, 96	96 = 32 × 3 + 0	32
1860, 2015	2015 = 1860 × 1 + 155 1860 = 155 × 12 + 0	155

Conclusion

Euclid's Division Algorithm provides a systematic and efficient method for finding the HCF of two positive integers. The practical activity with paper strips visually confirms that the HCF is the largest common length that can measure both given lengths without any remainder.

Experience of the Student

This project helped me understand that mathematics is not just about calculations but also about visual and practical understanding. Cutting the strips and physically measuring them made the concept of HCF very clear and interesting.

Acknowledgement

1. My sincere thanks to our Mathematics teacher for guiding us through this project

2. I would also like to thank my classmates for their cooperation and support

Reference Books / Resources

X Class State Board Mathematics Textbook
IX Class Mathematics Textbook

Mathsstudent.com

Project Work-2 Decoding Decimals: Terminating or Non-Terminating?

Name: R.Reema

Class: X

Roll Number: 10

Lesson: Real Numbers

Subject: Mathematics

Aim

To investigate the relationship between the prime factors of the denominator of a rational number and the nature of its decimal expansion (terminating or non-terminating repeating).

Objectives

Expressing rational numbers in their decimal form
Expressing the denominator of the rational number in its prime factorized form
Differentiating between terminating and non-terminating recurring decimals

Materials & Tools

Materials Used: Pen, Pencil, Scale, Chart Paper, Coloured Pens

Tools: Collection of rational numbers, Mathematical calculation, Prime factorization

Procedure

Collected various rational numbers
Each rational number was converted into its decimal form by performing division
Noted whether the decimal was Terminating (T) or Non-Terminating Repeating (NR)
Each rational number was written in its simplest form p/q
The denominator (q) was factorized into its prime factors
The observations were recorded in a table for analysis

Observations

Relation between Denominator and Decimal Expansion

S.No.	Rational Number	Decimal Form	Type	Simplified Form	Prime Factors of q	Observation
1	3/8	0.375	T	3/8	2³	Only prime factor 2
2	7/25	0.28	T	7/25	5²	Only prime factor 5
3	13/125	0.104	T	13/125	5³	Only prime factor 5
4	9/20	0.45	T	9/20	2² × 5¹	Prime factors 2 and 5
5	1/3	0.333...	NR	1/3	3¹	Prime factor 3
6	5/12	0.41666...	NR	5/12	2² × 3¹	Has prime factor 3
7	7/13	0.538461...	NR	7/13	13¹	Prime factor 13
8	29/343	0.08454...	NR	29/343	7³	Prime factor 7

Conclusion

From the observations, it is concluded that the decimal expansion of a rational number p/q (in simplest form) terminates if and only if the prime factorization of the denominator q is of the form 2ⁿ × 5ᵐ, where n and m are non-negative integers. If the denominator has any prime factor other than 2 or 5, the decimal expansion is non-terminating and repeating.

Result

A rational number p/q, where p and q are co-prime, has a terminating decimal expansion if the prime factors of q are only 2 and/or 5.

Experience of the Student

While working on this project, I learned to predict the nature of a decimal just by looking at the denominator of a fraction. Initially, I thought all decimals either stop or show a simple pattern, but I discovered that some repeats after a long sequence, which was fascinating.

Acknowledgement

1. I thank our Mathematics teacher for her invaluable guidance

2. I extend my gratitude to my parents for their support

Reference Books / Resources

X Class State Board Mathematics Textbook (Chapter 1: Real Numbers)
IX Class Mathematics Textbook

TS Board 10th Previous Questions -Solutions

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TS Board 10th Previous Questions - Solutions

Complete Chapter-wise Solutions for Telangana State Board Class 10 Mathematics

Access comprehensive solutions to previous years' questions for Telangana State Board Class 10 Mathematics. Each chapter includes detailed step-by-step solutions to help you understand concepts and excel in your exams.

Real Numbers

Explore the fundamental properties of real numbers, including Euclid's division algorithm, the Fundamental Theorem of Arithmetic, and irrational numbers.

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Sets

Understand set theory concepts including types of sets, set operations, Venn diagrams, and applications of sets in problem-solving.

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Polynomials

Master polynomial operations, zeroes of polynomials, relationship between zeroes and coefficients, and division algorithm for polynomials.

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Pair of Linear Equations in Two Variables

Solve systems of linear equations using graphical method, substitution, elimination, and cross-multiplication methods with practical applications.

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Quadratic Equations

Learn to solve quadratic equations by factorization, completing the square, and using the quadratic formula. Understand the nature of roots.

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Progressions

Explore arithmetic and geometric progressions, find nth terms, sum of n terms, and solve problems involving sequences and series.

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Coordinate Geometry

Understand the Cartesian plane, distance formula, section formula, and area of triangles using coordinate geometry concepts.

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Similar Triangles

Study similarity of triangles, criteria for similarity, basic proportionality theorem, and applications of similar triangles in problem-solving.

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Tangent and Secant to a Circle

Explore properties of tangents and secants to circles, theorems related to tangents, and solve problems involving circles and lines.

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Mensuration

Calculate surface areas and volumes of combinations of solids, conversion of solids from one shape to another, and frustum of a cone.

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Trigonometry

Learn trigonometric ratios, trigonometric identities, and solve problems involving heights and distances using trigonometry.

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Applications of Trigonometry

Apply trigonometric concepts to solve real-world problems involving heights and distances, angles of elevation and depression.

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Probability

Understand probability concepts, calculate probabilities of events, and solve problems using theoretical and experimental probability.

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Statistics

Learn to calculate mean, median, mode of grouped and ungrouped data, and understand graphical representation of data.

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Statistics – Solutions

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Statistics (1 Mark) - Complete Solutions

Important Statistics Concepts:

• Mean = Sum of all observations / Total number of observations
• Median = Middle value when data is arranged in order
• Mode = Most frequently occurring value
• Assumed Mean Method: A = Assumed mean, d = x - A
• Median for grouped data: l + [(n/2 - cf)/f] × h
• Outliers affect mean more than median or mode

Class Interval: 10–25, 25–40, 40–55, 55–70, 70–85, 85–100
Frequency: 2, 3, 7, 6, 6, 6
How do you find the deviation from the assumed mean for the above data? (M'15)

Step 1: Choose assumed mean (A)

Select a convenient value from the class marks (midpoints) as assumed mean.
Usually, we choose the class mark of the middle class.

Step 2: Find class marks

Class mark = (Lower limit + Upper limit) / 2
For 10-25: (10+25)/2 = 17.5
For 25-40: (25+40)/2 = 32.5
For 40-55: (40+55)/2 = 47.5
For 55-70: (55+70)/2 = 62.5
For 70-85: (70+85)/2 = 77.5
For 85-100: (85+100)/2 = 92.5

Step 3: Calculate deviation (d)

Let A = 47.5 (assumed mean)
d = Class mark - A
For each class: d = x - 47.5

∴ Deviation from assumed mean = Class mark - Assumed mean (d = x - A)

Write the formula to find the median of a grouped data and explain each term. (M'16, Aug'22, A'23)

Step 1: Median formula for grouped data

Median = l + [(n/2 - cf)/f] × h

Step 2: Explanation of terms

l = Lower limit of median class
n = Total number of observations
cf = Cumulative frequency of class preceding median class
f = Frequency of median class
h = Class width of median class

∴ Median = l + [(n/2 - cf)/f] × h, where l is lower limit, n is total frequency, cf is cumulative frequency before median class, f is frequency of median class, h is class width.

When an observation in a data is abnormally more than or less than the remaining observations in the data, does it affect the mean or mode or median? Why? (J'15)

Step 1: Understand the effect on mean

Mean is affected because it considers the value of every observation.
Mean = Sum of all observations / Total number of observations
An extreme value changes the sum significantly.

Step 2: Understand the effect on median

Median is less affected because it depends only on the position, not the values.
It is the middle value when data is arranged in order.

Step 3: Understand the effect on mode

Mode is least affected because it depends on frequency, not values.
Unless the extreme value repeats, it won't affect mode.

∴ An abnormal observation affects the mean the most because mean considers the actual values of all observations. Median and mode are less affected as they depend on position and frequency respectively.

Write the formula to find the mean of a grouped data, using assumed mean method and explain each term. (J'16)

Step 1: Assumed mean method formula

Mean = A + (Σfd/Σf)

Step 2: Explanation of terms

A = Assumed mean (conveniently chosen class mark)
d = Deviation = x - A (x is class mark)
f = Frequency of each class
Σfd = Sum of (frequency × deviation)
Σf = Total frequency = n

∴ Mean = A + (Σfd/Σf), where A is assumed mean, d is deviation (x - A), f is frequency, Σfd is sum of (f × d), and Σf is total frequency.

"The median of observations, –2, 5, 3, –1, 4, 6 is 3.5". Is it correct? (M'17)

Step 1: Arrange data in ascending order

Original data: -2, 5, 3, -1, 4, 6
Arranged in order: -2, -1, 3, 4, 5, 6

Step 2: Find median

Number of observations (n) = 6 (even)
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 3rd and 4th terms
= (3 + 4)/2 = 7/2 = 3.5

∴ Yes, the statement is correct. The median is indeed 3.5.

Write the first 10 prime numbers and find their median. (J'17)

Step 1: List first 10 prime numbers

Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Step 2: Find median

Number of observations (n) = 10 (even)
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 5th and 6th terms
= (11 + 13)/2 = 24/2 = 12

∴ The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and their median is 12.

Write the formula to find the median of grouped data and explain the alphabet in it. (M'18)

Step 1: Median formula for grouped data

Median = l + [(n/2 - cf)/f] × h

Step 2: Explanation of alphabets

l = Lower limit of median class
n = Total number of observations
cf = Cumulative frequency of class preceding median class
f = Frequency of median class
h = Class width of median class

∴ Median = l + [(n/2 - cf)/f] × h, where l is lower limit, n is total frequency, cf is cumulative frequency before median class, f is frequency of median class, h is class width.

Prathyusha stated that "the average of first 10 odd numbers is also 10". Do you agree with her? Justify your answer. (M'18)

Step 1: List first 10 odd numbers

First 10 odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Step 2: Calculate average

Sum = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
Average = Sum / 10 = 100 / 10 = 10

∴ Yes, I agree with Prathyusha. The average of first 10 odd numbers is indeed 10.

Find the median of first seven composite numbers. (M'19)

Step 1: List first seven composite numbers

Composite numbers: Numbers with more than 2 factors
First seven composite numbers: 4, 6, 8, 9, 10, 12, 14

Step 2: Find median

Number of observations (n) = 7 (odd)
Median = (n+1)/2 th term = 8/2 = 4th term
4th term = 9

∴ The median of first seven composite numbers is 9.

Find the mode of the data 6, 8, 3, 6, 3, 7, 4, 6, 7, 3, 6. (J'19)

Step 1: Count frequency of each number

6 appears 4 times
3 appears 3 times
7 appears 2 times
8 appears 1 time
4 appears 1 time

Step 2: Identify mode

Mode = Most frequently occurring value
6 occurs most frequently (4 times)

∴ The mode of the given data is 6.

Statistics (2 Marks) - Complete Solutions

Important Statistics Concepts:

• Mean = Sum of all observations / Total number of observations
• Median = Middle value when data is arranged in order
• Mode = Most frequently occurring value
• Mean for grouped data: Σfixi / Σfi
• Mode for grouped data: l + [(f1 - f0)/(2f1 - f0 - f2)] × h

The heights of six members of a family are given below in the table. Find the mean height of the family members. (J'15)

Step 1: Understand the data

Height (in ft.): 5, 5.2, 5.4, 5.6
Number of Family members: 1, 2, 2, 1

Step 2: Calculate total height

Total height = (5 × 1) + (5.2 × 2) + (5.4 × 2) + (5.6 × 1)
= 5 + 10.4 + 10.8 + 5.6 = 31.8 ft

Step 3: Calculate total family members

Total family members = 1 + 2 + 2 + 1 = 6

Step 4: Calculate mean height

Mean height = Total height / Total family members
= 31.8 / 6 = 5.3 ft

∴ The mean height of the family members is 5.3 ft

Find the value of Σfixi for the above data, where xi is the mid value of each class. (J'16)

Step 1: Understand the data

Class Intervals: 10-20, 20-30, 30-40, 40-50, 50-60
Frequencies (fi): 5, 8, 10, 5, 2

Step 2: Find mid values (xi)

For 10-20: xi = (10+20)/2 = 15
For 20-30: xi = (20+30)/2 = 25
For 30-40: xi = (30+40)/2 = 35
For 40-50: xi = (40+50)/2 = 45
For 50-60: xi = (50+60)/2 = 55

Step 3: Calculate Σfixi

Σfixi = (5×15) + (8×25) + (10×35) + (5×45) + (2×55)
= 75 + 200 + 350 + 225 + 110 = 960

∴ The value of Σfixi is 960

Find the median of 2/3, 4/5, 1/2, 3/4, 6/5 (M'18)

Step 1: Convert fractions to decimals for easier comparison

2/3 ≈ 0.6667
4/5 = 0.8
1/2 = 0.5
3/4 = 0.75
6/5 = 1.2

Step 2: Arrange in ascending order

Original: 2/3, 4/5, 1/2, 3/4, 6/5
In order: 1/2, 2/3, 3/4, 4/5, 6/5

Step 3: Find median

Number of observations (n) = 5 (odd)
Median = (n+1)/2 th term = 6/2 = 3rd term
3rd term = 3/4

∴ The median of the given fractions is 3/4

Find the mean of prime numbers less than 30. (J'18)

Step 1: List prime numbers less than 30

Prime numbers less than 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Step 2: Calculate sum

Sum = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129

Step 3: Calculate mean

Number of primes = 10
Mean = Sum / Number of primes = 129 / 10 = 12.9

∴ The mean of prime numbers less than 30 is 12.9

Write the mode formula for grouped data and explain the terms in it. (J'15, M'17,19,'22, J'23)

Step 1: Mode formula for grouped data

Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h

Step 2: Explanation of terms

l = Lower limit of modal class
f₁ = Frequency of modal class
f₀ = Frequency of class preceding modal class
f₂ = Frequency of class succeeding modal class
h = Class width of modal class

Step 3: Additional notes

• Modal class is the class with highest frequency
• This formula gives the exact mode for grouped data
• It's based on the principle that mode lies in the class with maximum frequency

∴ Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h, where l is lower limit of modal class, f₁ is frequency of modal class, f₀ is frequency of preceding class, f₂ is frequency of succeeding class, and h is class width.

Find the median of first 6 prime numbers. (J'19)

Step 1: List first 6 prime numbers

First 6 prime numbers: 2, 3, 5, 7, 11, 13

Step 2: Find median

Number of observations (n) = 6 (even)
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 3rd and 4th terms
= (5 + 7)/2 = 12/2 = 6

∴ The median of first 6 prime numbers is 6

Statistics (4 Marks) - Complete Solutions

Important Statistics Concepts:

• Mean = Σfixi / Σfi
• Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h
• Median = l + [(n/2 - cf)/f] × h
• Step Deviation Method: Mean = A + (Σfidi/Σfi) × h
• Ogive: Cumulative frequency curve

In a village, an enumerator has surveyed for 25 households. The size of the family (number of family members) and the number of families is tabulated as follows. Find the mode of the data. (M'15)

Size of family	1-3	3-5	5-7	7-9	9-11
No. of families	6	7	9	2	1

Step 1: Identify modal class

Highest frequency = 9
Modal class = 5-7

Step 2: Apply mode formula

Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h

l = 5 (lower limit of modal class)
f₁ = 9 (frequency of modal class)
f₀ = 7 (frequency of class preceding modal class)
f₂ = 2 (frequency of class succeeding modal class)
h = 2 (class width)

Step 3: Calculate mode

Mode = 5 + [(9 - 7)/(2×9 - 7 - 2)] × 2
= 5 + [2/(18 - 9)] × 2
= 5 + (2/9) × 2
= 5 + 4/9 = 5 + 0.44 = 5.44

∴ The mode of the data is 5.44 family members

Daily expenditure of 25 householders is given in the following table. Draw a "less than type" cumulative frequency Ogive curve for this data. (M'15)

Daily expenditure (Rs)	100-150	150-200	200-250	250-300	300-350
No. of households	4	5	12	2	2

Step 1: Create less than cumulative frequency table

Daily expenditure (less than)	150	200	250	300	350
Cumulative frequency	4	9	21	23	25

Step 2: Plot points for ogive

Points to plot: (150, 4), (200, 9), (250, 21), (300, 23), (350, 25)

Step 3: Draw ogive curve

Less Than Ogive Curve:

X-axis: Daily Expenditure (Rs)

Y-axis: Cumulative Frequency

Plot points: (150,4), (200,9), (250,21), (300,23), (350,25)

Join points with a smooth freehand curve

∴ The less than ogive curve is drawn by plotting points (150,4), (200,9), (250,21), (300,23), (350,25) and joining them with a smooth curve.

If the median of 60 observations given below is 28.5, then find the values of x and y. (J'15)

Class Interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	5	x	20	15	y	5

Step 1: Total frequency equation

Total frequency = 60
5 + x + 20 + 15 + y + 5 = 60
x + y + 45 = 60
x + y = 15 ............ (1)

Step 2: Identify median class

n = 60, n/2 = 30
Median = 28.5 (given)
Median class = 20-30 (since median 28.5 lies in this class)

Step 3: Apply median formula

Median = l + [(n/2 - cf)/f] × h

l = 20, n/2 = 30, f = 20, h = 10
cf = cumulative frequency before median class = 5 + x
28.5 = 20 + [(30 - (5 + x))/20] × 10
28.5 - 20 = [(25 - x)/20] × 10
8.5 = (25 - x)/2
17 = 25 - x
x = 25 - 17 = 8

Step 4: Find y using equation (1)

x + y = 15
8 + y = 15
y = 15 - 8 = 7

∴ The values are x = 8 and y = 7

The following distribution gives the daily profits (in rupees) earned by 50 shops in a locality. Convert the above distribution to a 'less than type' cumulative frequency distribution and draw its Ogive. (J'15)

Daily Profits (Rs.)	0-50	50-100	100-150	150-200	200-250	250-300
No. of shops	6	9	13	10	8	4

Step 1: Create less than cumulative frequency table

Daily profits (less than)	50	100	150	200	250	300
Cumulative frequency	6	15	28	38	46	50

Step 2: Plot points for ogive

Points to plot: (50, 6), (100, 15), (150, 28), (200, 38), (250, 46), (300, 50)

Step 3: Draw ogive curve

Less Than Ogive Curve:

X-axis: Daily Profits (Rs)

Y-axis: Cumulative Frequency

Plot points: (50,6), (100,15), (150,28), (200,38), (250,46), (300,50)

Join points with a smooth freehand curve

∴ The less than ogive is drawn by plotting points (50,6), (100,15), (150,28), (200,38), (250,46), (300,50) and joining them with a smooth curve.

Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers in the factory by using step-deviation method. (M'16)

Daily wages (Rs.)	200-250	250-300	300-350	350-400	400-450
No. of workers	6	8	14	10	12

Step 1: Create calculation table

Class	Frequency (f)	Midpoint (x)	d = (x - A)/h A=325, h=50	fd
200-250	6	225	-2	-12
250-300	8	275	-1	-8
300-350	14	325	0	0
350-400	10	375	1	10
400-450	12	425	2	24
Total	50			14

Step 2: Apply step-deviation formula

Mean = A + (Σfd/Σf) × h

A = 325, h = 50, Σfd = 14, Σf = 50
Mean = 325 + (14/50) × 50
= 325 + 14 = 339

∴ The mean daily wage of the workers is Rs. 339

Probability – Solutions

zaheerpashamd

Probability - 1 Mark Solutions

Important Probability Concepts:

• Probability = Number of favorable outcomes / Total number of outcomes
• 0 ≤ P(E) ≤ 1
• P(E) + P(not E) = 1
• Prime numbers: Numbers with exactly two factors (1 and itself)
• Composite numbers: Numbers with more than two factors
• Equally likely events: Events with same probability of occurrence

When a die is rolled once unbiased, what is the probability of getting a multiple of 3 out of possible outcomes? (M'15)

Step 1: Identify total outcomes

When a die is rolled, total possible outcomes = 6
Sample space = {1, 2, 3, 4, 5, 6}

Step 2: Identify favorable outcomes

Multiples of 3 from 1 to 6: {3, 6}
Number of favorable outcomes = 2

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(multiple of 3) = 2/6 = 1/3

∴ The probability of getting a multiple of 3 is 1/3

The probability of an event is always in between 0 and 1 why? (J'15)

Step 1: Understand probability definition

Probability = Number of favorable outcomes / Total number of outcomes

Step 2: Analyze numerator and denominator

- Number of favorable outcomes is always ≥ 0
- Number of favorable outcomes is always ≤ Total number of outcomes
- Total number of outcomes is always ≥ 1

Step 3: Establish the range

Minimum value: When favorable outcomes = 0
P(E) = 0/Total outcomes = 0

Maximum value: When favorable outcomes = Total outcomes
P(E) = Total outcomes/Total outcomes = 1

For any other case: 0 < P(E) < 1

∴ Probability always lies between 0 and 1 because favorable outcomes can't be negative and can't exceed total outcomes.

Find the probability of getting a sum of the numbers on them is 7, when two dice are rolled at a time. (M'16)

Step 1: Identify total outcomes

When two dice are rolled:
Total outcomes = 6 × 6 = 36

Step 2: Identify favorable outcomes

Pairs with sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Number of favorable outcomes = 6

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(sum = 7) = 6/36 = 1/6

∴ The probability of getting a sum of 7 is 1/6

Find the probability of getting a prime number, when a card drawn at random from the numbered cards from 1 to 25. (J'16)

Step 1: Identify total outcomes

Cards numbered from 1 to 25
Total outcomes = 25

Step 2: Identify favorable outcomes

Prime numbers from 1 to 25: 2, 3, 5, 7, 11, 13, 17, 19, 23
Number of favorable outcomes = 9

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(prime) = 9/25

∴ The probability of getting a prime number is 9/25

From the first 50 natural numbers, find the probability of randomly selected number is a multiple of 3. (M'17)

Step 1: Identify total outcomes

First 50 natural numbers: 1 to 50
Total outcomes = 50

Step 2: Identify favorable outcomes

Multiples of 3 from 1 to 50: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48
Number of favorable outcomes = 16

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(multiple of 3) = 16/50 = 8/25

∴ The probability of getting a multiple of 3 is 8/25

A dice is thrown once. Find the probability of getting a composite number. (J'17)

Step 1: Identify total outcomes

When a die is thrown, total possible outcomes = 6
Sample space = {1, 2, 3, 4, 5, 6}

Step 2: Identify favorable outcomes

Composite numbers from 1 to 6: 4, 6
(Note: 1 is neither prime nor composite)
Number of favorable outcomes = 2

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(composite) = 2/6 = 1/3

∴ The probability of getting a composite number is 1/3

What is the probability of getting exactly 2 heads, when three coins are tossed simultaneously. (M'18)

Step 1: Identify total outcomes

When three coins are tossed:
Total outcomes = 2 × 2 × 2 = 8
Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Step 2: Identify favorable outcomes

Outcomes with exactly 2 heads: HHT, HTH, THH
Number of favorable outcomes = 3

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(exactly 2 heads) = 3/8

∴ The probability of getting exactly 2 heads is 3/8

When a dice is rolled, find the probability of getting an odd prime number. (J'18)

Step 1: Identify total outcomes

When a die is rolled, total possible outcomes = 6
Sample space = {1, 2, 3, 4, 5, 6}

Step 2: Identify favorable outcomes

Odd prime numbers from 1 to 6: 3, 5
(Note: 2 is prime but not odd)
Number of favorable outcomes = 2

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(odd prime) = 2/6 = 1/3

∴ The probability of getting an odd prime number is 1/3

From English alphabet if a letter is chosen at random, then find the probability that the letter is a consonant. (M'19)

Step 1: Identify total outcomes

Total letters in English alphabet = 26

Step 2: Identify favorable outcomes

Vowels: A, E, I, O, U (5 vowels)
Consonants = Total letters - Vowels = 26 - 5 = 21

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(consonant) = 21/26

∴ The probability of getting a consonant is 21/26

Write two examples for equally likely events. (J'19)

Step 1: Understand equally likely events

Equally likely events are events that have the same probability of occurrence.

Step 2: Example 1

When a fair coin is tossed:
P(Head) = 1/2 and P(Tail) = 1/2
Getting a Head and getting a Tail are equally likely events.

Step 3: Example 2

When a fair die is rolled:
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6
Getting any specific number from 1 to 6 are equally likely events.

∴ Two examples of equally likely events are: (1) Getting Head or Tail when tossing a fair coin, (2) Getting any number from 1 to 6 when rolling a fair die.

If a dice rolled once, then find the probability of getting an odd number. (May 2022)

Step 1: Identify total outcomes

When a die is rolled, total possible outcomes = 6
Sample space = {1, 2, 3, 4, 5, 6}

Step 2: Identify favorable outcomes

Odd numbers from 1 to 6: 1, 3, 5
Number of favorable outcomes = 3

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(odd) = 3/6 = 1/2

∴ The probability of getting an odd number is 1/2

If an unbiased dice is rolled once, then find the probability of getting a prime number on its top face. (Aug 22)

Step 1: Identify total outcomes

When a die is rolled, total possible outcomes = 6
Sample space = {1, 2, 3, 4, 5, 6}

Step 2: Identify favorable outcomes

Prime numbers from 1 to 6: 2, 3, 5
(Note: 1 is not a prime number)
Number of favorable outcomes = 3

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(prime) = 3/6 = 1/2

∴ The probability of getting a prime number is 1/2

Find the probability of getting a 'vowel' if a letter is chosen randomly from the word "INNOVATION". (Apr'23)

Step 1: Identify total outcomes

Letters in "INNOVATION": I, N, N, O, V, A, T, I, O, N
Total letters = 10

Step 2: Identify favorable outcomes

Vowels in "INNOVATION": I, O, A, I, O
Number of vowels = 5

Step 3: Apply probability formula

Probability = Favorable outcomes / Total outcomes
P(vowel) = 5/10 = 1/2

∴ The probability of getting a vowel is 1/2

Probability (2 Marks) - Complete Solutions

Important Probability Concepts:

• Probability = Number of favorable outcomes / Total number of outcomes
• P(not E) = 1 - P(E)
• Composite numbers: Numbers with more than two factors
• Prime numbers: Numbers with exactly two factors (1 and itself)
• Face cards: Jack, Queen, King (12 in a deck)
• Aces: 4 in a deck

There are 12 red balls, 18 blue balls and 6 white balls in a box. When a ball is drawn at random from the box, what is the probability of not getting a red ball? (M'15)

Step 1: Find total number of balls

Red balls = 12
Blue balls = 18
White balls = 6
Total balls = 12 + 18 + 6 = 36

Step 2: Find probability of getting a red ball

P(red) = Number of red balls / Total balls = 12/36 = 1/3

Step 3: Find probability of not getting a red ball

P(not red) = 1 - P(red) = 1 - 1/3 = 2/3

∴ The probability of not getting a red ball is 2/3

When a card is drawn from a well shuffled deck of 52 cards, then find the probability of NOT getting a red faced card. (J'15)

Step 1: Understand the deck composition

Total cards = 52
Face cards: Jack, Queen, King (3 per suit)
Red suits: Hearts and Diamonds (2 suits)
Red face cards = 3 × 2 = 6

Step 2: Find probability of getting a red face card

P(red face card) = 6/52 = 3/26

Step 3: Find probability of not getting a red face card

P(not red face card) = 1 - P(red face card) = 1 - 3/26 = 23/26

∴ The probability of not getting a red face card is 23/26

There are 5 red balls, 4 green balls and 6 yellow balls in a box. If a ball is selected at random, what is the probability of not getting a yellow ball? (J'16)

Step 1: Find total number of balls

Red balls = 5
Green balls = 4
Yellow balls = 6
Total balls = 5 + 4 + 6 = 15

Step 2: Find probability of getting a yellow ball

P(yellow) = Number of yellow balls / Total balls = 6/15 = 2/5

Step 3: Find probability of not getting a yellow ball

P(not yellow) = 1 - P(yellow) = 1 - 2/5 = 3/5

∴ The probability of not getting a yellow ball is 3/5

What is the probability of a number picked from first 20 natural numbers is even composite number? (M'18)

Step 1: Identify total outcomes

First 20 natural numbers: 1 to 20
Total outcomes = 20

Step 2: Identify favorable outcomes

Even composite numbers from 1 to 20: 4, 6, 8, 10, 12, 14, 16, 18, 20
(Note: 2 is prime, not composite)
Number of favorable outcomes = 9

Step 3: Apply probability formula

P(even composite) = 9/20

∴ The probability of getting an even composite number is 9/20

A bag contains 7 red, 5 white and 6 black balls. A ball is drawn from the bag at random; find the probability that the ball drawn is not black. (J'18)

Step 1: Find total number of balls

Red balls = 7
White balls = 5
Black balls = 6
Total balls = 7 + 5 + 6 = 18

Step 2: Find probability of getting a black ball

P(black) = Number of black balls / Total balls = 6/18 = 1/3

Step 3: Find probability of not getting a black ball

P(not black) = 1 - P(black) = 1 - 1/3 = 2/3

∴ The probability that the ball drawn is not black is 2/3

A bag contains balls which are numbered from 1 to 50. A ball is drawn at random from the bag, the probability that it bears two digit number multiple of 7. (M'19)

Step 1: Identify total outcomes

Numbers from 1 to 50
Total outcomes = 50

Step 2: Identify favorable outcomes

Two-digit multiples of 7: 14, 21, 28, 35, 42, 49
Number of favorable outcomes = 6

Step 3: Apply probability formula

P(two-digit multiple of 7) = 6/50 = 3/25

∴ The probability of getting a two-digit number multiple of 7 is 3/25

A box contains 4 red balls, 5 green balls and P white balls. If the probability of randomly picked a ball from the box to be red ball is 1/3, then find the number of white balls. (J'19)

Step 1: Set up the equation

Red balls = 4
Green balls = 5
White balls = P
Total balls = 4 + 5 + P = 9 + P
P(red) = 4/(9 + P) = 1/3

Step 2: Solve for P

4/(9 + P) = 1/3
Cross multiply: 4 × 3 = 1 × (9 + P)
12 = 9 + P
P = 12 - 9 = 3

∴ The number of white balls is 3

A bag contains 5 red, 8 white, 4 green colour balls. If a ball is selected randomly from the bag then find the probability that selected ball is (i) a green ball (ii) not white ball.

Step 1: Find total number of balls

Red balls = 5
White balls = 8
Green balls = 4
Total balls = 5 + 8 + 4 = 17

(i) Probability of green ball

P(green) = Number of green balls / Total balls = 4/17

(ii) Probability of not white ball

P(white) = 8/17
P(not white) = 1 - P(white) = 1 - 8/17 = 9/17

∴ (i) P(green) = 4/17, (ii) P(not white) = 9/17

A box contains four slips numbered 1, 2, 3, 4 and another box contains five slips numbered 5, 6, 7, 8, 9. If one slip is taken randomly from each box, i) How many number pairs are possible? ii) What is the probability of both being odd? iii) What is the probability of getting the sum of the numbers 10? (Apr'23)

Step 1: Total possible pairs

First box: 4 slips
Second box: 5 slips
Total pairs = 4 × 5 = 20

(i) Number of possible pairs

Total pairs = 20

(ii) Probability of both being odd

Odd numbers in first box: 1, 3 (2 numbers)
Odd numbers in second box: 5, 7, 9 (3 numbers)
Favorable pairs = 2 × 3 = 6
P(both odd) = 6/20 = 3/10

(iii) Probability of sum being 10

Pairs with sum 10: (1,9), (2,8), (3,7), (4,6)
Number of favorable pairs = 4
P(sum = 10) = 4/20 = 1/5

∴ (i) 20 pairs, (ii) P(both odd) = 3/10, (iii) P(sum = 10) = 1/5

If one card is randomly selected from a well shuffled deck of cards, then find the probability of getting- i) a face card, (ii) a jack of hearts and (iii) an ace card. (Jun'23)

Step 1: Understand deck composition

Total cards = 52
Face cards: Jack, Queen, King (3 per suit × 4 suits = 12)
Aces: 1 per suit × 4 suits = 4

(i) Probability of face card

P(face card) = 12/52 = 3/13

(ii) Probability of jack of hearts

There is only 1 jack of hearts
P(jack of hearts) = 1/52

(iii) Probability of ace card

P(ace) = 4/52 = 1/13

∴ (i) P(face card) = 3/13, (ii) P(jack of hearts) = 1/52, (iii) P(ace) = 1/13

Probability (4 Marks) - Complete Solutions

Important Probability Concepts:

• Probability = Number of favorable outcomes / Total number of outcomes
• P(not E) = 1 - P(E)
• Prime numbers: Numbers with exactly two factors (1 and itself)
• Composite numbers: Numbers with more than two factors
• Face cards: Jack, Queen, King (12 in a deck)
• Aces: 4 in a deck

There are 100 flash cards labeled from 1 to 100 in a bag. When a card is drawn from the bag at random, what is the probability of getting…… i) A card with prime number from possible outcomes ii) A card without prime number from possible outcomes. (M'15)

Step 1: Find total outcomes

Total cards = 100

(i) Probability of prime number

Step 2: Find prime numbers between 1 and 100

Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Total prime numbers = 25

Step 3: Calculate probability

P(prime) = 25/100 = 1/4

(ii) Probability of not prime number

Step 4: Calculate probability

P(not prime) = 1 - P(prime) = 1 - 1/4 = 3/4
OR
Non-prime numbers = 100 - 25 = 75
P(not prime) = 75/100 = 3/4

∴ (i) P(prime) = 1/4, (ii) P(not prime) = 3/4

A shopkeeper has 100 memory cards in a box. Among them, 15 memory cards are defective. When a person came to the shop to buy a memory card, the shopkeeper drew a memory card at random from the box. Then What is the probability that this memory card is defective? After drawing the first memory card which is defective, it is not placed back in the box. Then another memory card is drawn at random. What is the probability that this memory card is NOT defective? (J'15)

(i) Probability first card is defective

Step 1: Find total outcomes

Total cards = 100
Defective cards = 15
P(defective) = 15/100 = 3/20

(ii) Probability second card is not defective

Step 2: Update counts after first draw

First card drawn is defective
Remaining cards = 100 - 1 = 99
Remaining defective cards = 15 - 1 = 14
Non-defective cards = 99 - 14 = 85

Step 3: Calculate probability

P(second card not defective) = 85/99

∴ (i) P(defective) = 3/20, (ii) P(second card not defective) = 85/99

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of blue balls in the bag. (M'16)

Step 1: Define variables

Let number of blue balls = x
Total balls = 5 + x

Step 2: Write probability expressions

P(red) = 5/(5 + x)
P(blue) = x/(5 + x)

Step 3: Set up equation

Given: P(blue) = 2 × P(red)
x/(5 + x) = 2 × 5/(5 + x)
x/(5 + x) = 10/(5 + x)

Step 4: Solve for x

x = 10 (since denominators are equal and non-zero)

∴ The number of blue balls is 10

Two dice are rolled at same time and the sum of the numbers appearing on them is noted. Find the probability of getting each sum, from 3 to 5 separately. (J'16)

Step 1: Find total outcomes

When two dice are rolled:
Total outcomes = 6 × 6 = 36

Probability of sum = 3

Step 2: Find favorable outcomes for sum 3

Pairs with sum 3: (1,2), (2,1)
Number of favorable outcomes = 2
P(sum = 3) = 2/36 = 1/18

Probability of sum = 4

Step 3: Find favorable outcomes for sum 4

Pairs with sum 4: (1,3), (2,2), (3,1)
Number of favorable outcomes = 3
P(sum = 4) = 3/36 = 1/12

Probability of sum = 5

Step 4: Find favorable outcomes for sum 5

Pairs with sum 5: (1,4), (2,3), (3,2), (4,1)
Number of favorable outcomes = 4
P(sum = 5) = 4/36 = 1/9

∴ P(sum = 3) = 1/18, P(sum = 4) = 1/12, P(sum = 5) = 1/9

A bag contains some square cards. A prime number between 1 and 100 has been written on each card. Find the probability of getting a card that the sum of the digits of a prime number written on it, is 8. (M'17)

Step 1: Find total prime numbers between 1 and 100

Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Total prime numbers = 25

Step 2: Find primes with digit sum = 8

Check each prime:
17 → 1 + 7 = 8 ✓
53 → 5 + 3 = 8 ✓
71 → 7 + 1 = 8 ✓
No other primes between 1-100 have digit sum 8
Favorable outcomes = 3

Step 3: Calculate probability

P(digit sum = 8) = 3/25

∴ The probability of getting a card with prime number having digit sum 8 is 3/25

From the Deck of 52 cards, if a card is randomly chosen, find the probability of getting a card with (i) a prime number on it, (ii) face on it. (M'18)

Step 1: Understand deck composition

Total cards = 52
Number cards: 2, 3, 4, 5, 6, 7, 8, 9, 10 (9 per suit × 4 suits = 36)
Face cards: Jack, Queen, King (3 per suit × 4 suits = 12)
Aces: 1 per suit × 4 suits = 4

(i) Probability of prime number card

Step 2: Find prime number cards

Prime numbers: 2, 3, 5, 7 (4 per suit)
Total prime number cards = 4 × 4 = 16
P(prime) = 16/52 = 4/13

(ii) Probability of face card

Step 3: Calculate probability

Face cards = 12
P(face) = 12/52 = 3/13

∴ (i) P(prime) = 4/13, (ii) P(face) = 3/13

If two dice are thrown at the same time, find the probability of getting sum of the dots on top is prime. (M'19)

Step 1: Find total outcomes

When two dice are thrown:
Total outcomes = 6 × 6 = 36

Step 2: Find prime sums possible

Possible sums: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Prime sums: 2, 3, 5, 7, 11

Step 3: Find favorable outcomes for each prime sum

Sum = 2: (1,1) → 1 outcome
Sum = 3: (1,2), (2,1) → 2 outcomes
Sum = 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes
Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
Sum = 11: (5,6), (6,5) → 2 outcomes
Total favorable outcomes = 1 + 2 + 4 + 6 + 2 = 15

Step 4: Calculate probability

P(prime sum) = 15/36 = 5/12

∴ The probability of getting a prime sum is 5/12

From a pack of 52 playing cards, Jacks, Queens, Kings and Aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is (i) a black queen, (ii) a red card. (J'19)

Step 1: Find cards removed

Red Jacks = 2, Red Queens = 2, Red Kings = 2, Red Aces = 2
Total cards removed = 8
Cards remaining = 52 - 8 = 44

(i) Probability of black queen

Step 2: Find black queens remaining

Black queens = 2 (not removed)
P(black queen) = 2/44 = 1/22

(ii) Probability of red card

Step 3: Find red cards remaining

Original red cards = 26
Red cards removed = 8
Red cards remaining = 26 - 8 = 18
P(red card) = 18/44 = 9/22

∴ (i) P(black queen) = 1/22, (ii) P(red card) = 9/22

A box contains 20 cards which are numbered from 1 to 20. If one card is selected at random from the box, find the probability that it bears (i) a prime number, (ii) an even number. (May 2022)

Step 1: Find total outcomes

Total cards = 20

(i) Probability of prime number

Step 2: Find prime numbers between 1 and 20

Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19
Total prime numbers = 8
P(prime) = 8/20 = 2/5

(ii) Probability of even number

Step 3: Find even numbers between 1 and 20

Even numbers: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Total even numbers = 10
P(even) = 10/20 = 1/2

∴ (i) P(prime) = 2/5, (ii) P(even) = 1/2

Applications of Trigonometry – Solutions

zaheerpashamd

Applications of Trigonometry - 1 Mark Solutions

Important Trigonometric Concepts:

• Angle of Elevation: Angle between horizontal line and line of sight when looking up
• Angle of Depression: Angle between horizontal line and line of sight when looking down
• tan θ = Opposite/Adjacent
• Height = Distance × tan(angle of elevation)
• When object and shadow are equal, angle = 45°
• As angle increases from 0° to 90°, tan θ increases from 0 to ∞

A person from the top of a building of height 25m has observed another building's top and bottom at an angle of elevation of 45° and at an angle of depression 60° respectively. Draw a diagram for this data. (M'15)

Step 1: Understand the scenario

- First building height: 25m
- Observer at top of first building
- Angle of elevation to top of second building: 45°
- Angle of depression to bottom of second building: 60°

Step 2: Diagram description

The diagram shows:
- Two vertical buildings with some distance between them
- First building of height 25m
- Observer at top of first building
- Line of sight to top of second building at 45° (angle of elevation)
- Line of sight to bottom of second building at 60° (angle of depression)
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer at 25m height with lines of sight at 45° elevation and 60° depression.

A person observed the top of a tree at an angle of elevation of 60° when the observation point was 5m away from the foot of the tree. Draw a diagram for this data. (J'15)

Step 1: Understand the scenario

- Distance from tree: 5m
- Angle of elevation: 60°
- Observer at ground level

Step 2: Diagram description

The diagram shows:
- A vertical tree
- Observer at ground level, 5m from tree
- Line of sight to top of tree at 60° angle
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer 5m from tree with 60° angle of elevation to the top.

"If the angle of elevation of Sun increases from 0° to 90°, then the length of a tower decreases", is this statement true? Justify your answer. (M'16)

Step 1: Understand the relationship

Let:
- h = height of tower (constant)
- θ = angle of elevation of sun
- l = length of shadow

Step 2: Apply trigonometry

tan θ = h/l
So, l = h / tan θ = h × cot θ

Step 3: Analyze the behavior

As θ increases from 0° to 90°:
- tan θ increases from 0 to ∞
- cot θ decreases from ∞ to 0
Therefore, l = h × cot θ decreases from ∞ to 0

Shadow Length Calculator

Tower Height (m):

Sun Angle (degrees):

∴ Yes, the statement is true. As the sun's angle increases from 0° to 90°, the shadow length decreases from infinity to zero.

If a tower of height 'h' is observed from a point with a distance 'd' and angle 'θ', then express the relation among h, d and θ. (J'16)

Step 1: Visualize the scenario

- Tower height = h
- Distance from tower = d
- Angle of elevation = θ

Step 2: Apply trigonometric ratio

In the right triangle formed:
- Opposite side = h (tower height)
- Adjacent side = d (distance)
- Angle = θ

Step 3: Write the relation

tan θ = Opposite/Adjacent = h/d
Therefore, h = d × tan θ

Diagram:

∴ The relation is h = d × tan θ

A pole and its shadow have same length, find the angle of the ray made with the earth at that time. (M'18)

Step 1: Understand the scenario

- Pole height = h
- Shadow length = h (same as pole height)
- We need to find the angle θ that the sun's rays make with the ground

Step 2: Apply trigonometry

In the right triangle formed:
- Opposite side = h (pole height)
- Adjacent side = h (shadow length)
tan θ = Opposite/Adjacent = h/h = 1

Step 3: Find the angle

tan θ = 1
We know that tan 45° = 1
Therefore, θ = 45°

Diagram:

∴ The angle of the sun's rays with the earth is 45°.

An observer observes the top of a tower from two points on the same side of a tower and on the same straight line which are at distances of 5 m and 8 m with angles of elevation 60° and 45° respectively. Draw a suitable diagram for the given data. (Aug.22)

Step 1: Understand the scenario

- Tower of unknown height
- First observation point: 5m from tower, angle = 60°
- Second observation point: 8m from tower, angle = 45°
- Both points on same side of tower and in straight line

Step 2: Diagram description

The diagram shows:
- A vertical tower
- Two observation points at distances 5m and 8m from tower
- Lines of sight from both points to top of tower
- Angles of elevation marked as 60° and 45°

Diagram:

∴ The diagram shows two observation points at 5m and 8m with elevation angles 60° and 45° respectively.

"An observer standing at a distance of 10m from the foot of a tower, observes its top with an angle of elevation of 60°". Draw a suitable diagram for this situation. (Apr'.23)

Step 1: Understand the scenario

- Distance from tower: 10m
- Angle of elevation: 60°
- Observer at ground level

Step 2: Diagram description

The diagram shows:
- A vertical tower
- Observer at ground level, 10m from tower
- Line of sight to top of tower at 60° angle
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer 10m from tower with 60° angle of elevation to the top.

Applications of Trigonometry (2 Marks) - Complete Solutions

Important Trigonometric Concepts:

A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to stream of the river and travels a distance of 450m to reach the another side of the river. Draw the diagram for this data. (M'16)

Step 1: Understand the scenario

- Boat makes 60° angle with river bank
- Distance traveled: 450m
- Due to stream, boat doesn't travel perpendicular to bank

Step 2: Diagram description

The diagram shows:
- Two parallel lines representing river banks
- Boat starting from one bank at 60° angle
- Actual path of boat (450m at 60°)
- Perpendicular distance between banks (width of river)
- Components: actual path and perpendicular crossing

Diagram:

∴ The diagram shows the boat crossing at 60° angle due to river stream.

A person 25 m away from a cell tower observes the top of the cell tower at an angle of elevation 30°. Draw the suitable diagram for this situation. (M'17)

Step 1: Understand the scenario

- Distance from tower: 25m
- Angle of elevation: 30°
- Observer at ground level

Step 2: Diagram description

The diagram shows:
- A vertical cell tower
- Observer at ground level, 25m from tower
- Line of sight to top of tower at 30° angle
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer 25m from cell tower with 30° angle of elevation.

A State highway leads to foot of the tower. A Man Standing at the top of the tower observes a car at an angle of depression of θ, which is approaching to the foot of the tower with a uniform speed. 6 seconds later the angle of depression is φ. Draw a diagram for this data and analyze. (J'17)

Step 1: Understand the scenario

- Tower of height h
- Car moving towards tower on highway
- Initial angle of depression: θ
- After 6 seconds, angle of depression: φ
- φ > θ (as car gets closer, angle increases)
- Car moving with uniform speed

Step 2: Diagram description

The diagram shows:
- A vertical tower
- Highway leading to foot of tower
- Observer at top of tower
- Two positions of car: initial and after 6 seconds
- Lines of sight with angles θ and φ
- Distances from tower: d₁ and d₂ (d₁ > d₂)

Diagram:

Step 3: Analysis

Let:
- Tower height = h
- Initial distance = d₁
- Final distance = d₂
- Speed of car = v

From trigonometry:
tan θ = h/d₁ ⇒ d₁ = h/tan θ
tan φ = h/d₂ ⇒ d₂ = h/tan φ

Distance covered in 6 seconds = d₁ - d₂ = h(1/tan θ - 1/tan φ)
Speed v = (d₁ - d₂)/6 = h(1/tan θ - 1/tan φ)/6

∴ The diagram shows car approaching tower with angles θ and φ. Speed can be calculated using tower height and angles.

From the top of the tower of height h m height, Anusha observes the angles of depression of two points X and Y on the same side of the tower on the ground to be α and β. Draw the suitable figure for the given information. (M'18)

Step 1: Understand the scenario

- Tower height = h
- Two points X and Y on same side of tower
- Angles of depression: α and β
- Assume α > β (point X is closer than Y)

Step 2: Diagram description

The diagram shows:
- A vertical tower of height h
- Observer at top of tower
- Two points X and Y on ground
- Lines of sight to X and Y with angles α and β
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows Anusha observing points X and Y with angles of depression α and β.

The angle of elevation of the top of a tower from a point on the ground, which is 50m away from the foot of the tower, is 45°. Draw the diagram for the situation. (J'18)

Step 1: Understand the scenario

- Distance from tower: 50m
- Angle of elevation: 45°
- Observer at ground level

Step 2: Diagram description

The diagram shows:
- A vertical tower
- Observer at ground level, 50m from tower
- Line of sight to top of tower at 45° angle
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer 50m from tower with 45° angle of elevation.

From the top of the building the angle of elevation of the top of the cell tower is 60° and the angle of depression to its foot is 45°, if the distance of the building from the tower is 30m, draw the suitable diagram to the given data. (M'19)

Step 1: Understand the scenario

- Distance between building and tower: 30m
- Observer at top of building
- Angle of elevation to top of tower: 60°
- Angle of depression to foot of tower: 45°

Step 2: Diagram description

The diagram shows:
- Two vertical structures: building and cell tower
- Distance between them: 30m
- Observer at top of building
- Line of sight to top of tower at 60° (elevation)
- Line of sight to foot of tower at 45° (depression)
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows observer on building with 60° elevation to tower top and 45° depression to tower foot.

From the top of the building, The angle of elevation of the top of a TV tower is α° and the angle of depression to its (T.V. tower) foot is β°. If distance of the building from the tower is 'd' metres, draw the suitable diagram of the given data. (J'19)

Step 1: Understand the scenario

- Distance between building and tower: d meters
- Observer at top of building
- Angle of elevation to top of TV tower: α°
- Angle of depression to foot of TV tower: β°

Step 2: Diagram description

The diagram shows:
- Two vertical structures: building and TV tower
- Distance between them: d meters
- Observer at top of building
- Line of sight to top of TV tower at α° (elevation)
- Line of sight to foot of TV tower at β° (depression)
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows observer on building with α° elevation to TV tower top and β° depression to tower foot.

The statue stands on the top of 3 m tall pedestal. From a point on the ground angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the statue. (Aug.22)

Step 1: Understand the scenario

- Pedestal height = 3m
- Statue height = ? (let's call it h)
- Same observation point for both measurements
- Angle to top of statue = 60°
- Angle to top of pedestal = 45°

Step 2: Diagram description

The diagram shows:
- Pedestal of height 3m
- Statue of height h on top of pedestal
- Observer at ground level
- Lines of sight to top of pedestal (45°) and top of statue (60°)
- Distance from observer to pedestal: d

Diagram:

Step 3: Set up equations

Let:
- Distance from observer = d
- Pedestal height = 3m
- Statue height = h

For pedestal top (45°):
tan 45° = 3/d ⇒ 1 = 3/d ⇒ d = 3m

For statue top (60°):
tan 60° = (3 + h)/d ⇒ √3 = (3 + h)/3

Step 4: Solve for statue height

√3 = (3 + h)/3
3√3 = 3 + h
h = 3√3 - 3
h = 3(√3 - 1)

Using √3 ≈ 1.732:
h = 3(1.732 - 1) = 3 × 0.732 = 2.196m ≈ 2.2m

Calculate Statue Height

Pedestal Height (m):

Angle to Pedestal (°):

Angle to Statue (°):

∴ The height of the statue is 3(√3 - 1) m ≈ 2.2 m.

Applications of Trigonometry - 4 Marks Solutions

Important Trigonometric Concepts:

An observer flying in an altitude of 900m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships. (M'15)

Step 1: Understand the scenario

- Observer's altitude: 900m
- Angle of depression to first ship: 60°
- Angle of depression to second ship: 30°
- Both ships are in the same direction

Step 2: Apply trigonometry

Let the distance from observer's vertical to first ship = d₁
Let the distance from observer's vertical to second ship = d₂
tan(60°) = 900/d₁ ⇒ d₁ = 900/tan(60°) = 900/√3 = 300√3 m
tan(30°) = 900/d₂ ⇒ d₂ = 900/tan(30°) = 900/(1/√3) = 900√3 m

Step 3: Find distance between ships

Distance between ships = d₂ - d₁ = 900√3 - 300√3 = 600√3 m

∴ The distance between the two ships is 600√3 m ≈ 1039.23 m

A person from the top of a building of height 15 meters observes the top and the bottom of a cell tower with the angle of elevation as 60° and the angle of depression as 45° respectively. Then find the height of the cell tower. (J'15)

Step 1: Understand the scenario

- Building height: 15m
- Angle of elevation to top of tower: 60°
- Angle of depression to bottom of tower: 45°
- We need to find height of cell tower

Step 2: Apply trigonometry

Let distance between buildings = d
From angle of depression 45°: tan(45°) = 15/d ⇒ d = 15m
Let height of tower above observer = h
From angle of elevation 60°: tan(60°) = h/d ⇒ h = d × tan(60°) = 15 × √3 = 15√3 m

Step 3: Find total height of tower

Total height of tower = height below observer + height above observer
= 15 + 15√3 = 15(1 + √3) m

∴ The height of the cell tower is 15(1 + √3) m ≈ 40.98 m

Two poles of equal heights are standing opposite to each other, on either side of the road, which is 80m wide. From a point between them on the road, the angles of elevation of top of the poles are 60° and 30° respectively. Find the height of the poles. (M'16)

Step 1: Understand the scenario

- Road width: 80m
- Two poles of equal height h
- Angles of elevation: 60° and 30°
- Observation point is between them on the road

Step 2: Apply trigonometry

Let distance from observation point to first pole = x
Then distance to second pole = 80 - x
For first pole: tan(60°) = h/x ⇒ h = x√3
For second pole: tan(30°) = h/(80 - x) ⇒ h = (80 - x)/√3

Step 3: Solve for x and h

Equating both expressions for h:
x√3 = (80 - x)/√3
3x = 80 - x
4x = 80 ⇒ x = 20m
Then h = x√3 = 20√3 m

∴ The height of each pole is 20√3 m ≈ 34.64 m

A tree is broken without separating from the stem by the wind. The top touches the ground making an angle 30° at a distance of 12m from the foot of the tree. Find the height of the tree before breaking. (J'16)

Step 1: Understand the scenario

- Tree broken but still attached to stem
- Top touches ground at 30° angle
- Distance from foot to where top touches: 12m
- Need to find original height of tree

Step 2: Apply trigonometry

Let height of standing part = h₁
Let length of broken part = h₂
In the triangle formed by broken part:
cos(30°) = 12/h₂ ⇒ h₂ = 12/cos(30°) = 12/(√3/2) = 24/√3 = 8√3 m
sin(30°) = h₁/12 ⇒ h₁ = 12 × sin(30°) = 12 × 1/2 = 6m

Step 3: Find total height

Total height = h₁ + h₂ = 6 + 8√3 m

∴ The original height of the tree was 6 + 8√3 m ≈ 19.86 m

Two poles are standing opposite to each other on the either side of the road which is 90 feet wide. The angle of elevation from bottom of the first pole to the top of the second pole is 45°. The angle of elevation from the bottom of the second pole to the top of the first pole is 30°. Find the heights of the poles. (use √3 = 1.732) (M'17)

Step 1: Understand the scenario

- Road width: 90 feet
- Two poles of different heights h₁ and h₂
- Angle from bottom of first to top of second: 45°
- Angle from bottom of second to top of first: 30°

Step 2: Apply trigonometry

For first observation: tan(45°) = h₂/90 ⇒ h₂ = 90 × 1 = 90 feet
For second observation: tan(30°) = h₁/90 ⇒ h₁ = 90 × 1/√3 = 90/√3 = 30√3 feet

Step 3: Calculate numerical values

h₁ = 30√3 ≈ 30 × 1.732 = 51.96 feet
h₂ = 90 feet

∴ The heights of the poles are approximately 51.96 feet and 90 feet

The angle of elevation of top of the tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it, are complementary. Prove that the height of the tower is 6 m. (J'17)

Step 1: Understand the scenario

- Tower height: h
- Two observation points at 4m and 9m from base
- Angles of elevation: θ and (90° - θ) [complementary]
- Need to prove h = 6m

Step 2: Apply trigonometry

From first point: tan θ = h/4
From second point: tan(90° - θ) = h/9 ⇒ cot θ = h/9
Since tan θ × cot θ = 1
(h/4) × (h/9) = 1
h²/36 = 1 ⇒ h² = 36 ⇒ h = 6m

∴ The height of the tower is 6 m, as required to prove.

From the top of a tower of 50m high, Neha observes the angles of depression of the top and foot of another building to be 45° and 60° respectively. Find the height of the building. (M'18)

Step 1: Understand the scenario

- First tower height: 50m
- Angle of depression to top of building: 45°
- Angle of depression to foot of building: 60°
- Need to find height of building

Step 2: Apply trigonometry

Let distance between buildings = d
From angle of depression to foot: tan(60°) = 50/d ⇒ d = 50/√3 m
Let height difference between towers = h
From angle of depression to top: tan(45°) = h/d ⇒ h = d = 50/√3 m

Step 3: Find building height

Building height = 50 - h = 50 - 50/√3 = 50(1 - 1/√3) m

∴ The height of the building is 50(1 - 1/√3) m ≈ 21.13 m

Two boys on either side of their school building of 20 m height observes its top at the angles of elevation 30° and 60° respectively. Find the distance between two boys. (J'18)

Step 1: Understand the scenario

- Building height: 20m
- Angles of elevation: 30° and 60°
- Boys on opposite sides of building
- Need to find distance between boys

Step 2: Apply trigonometry

For first boy: tan(30°) = 20/d₁ ⇒ d₁ = 20/tan(30°) = 20√3 m
For second boy: tan(60°) = 20/d₂ ⇒ d₂ = 20/tan(60°) = 20/√3 m

Step 3: Find total distance

Distance between boys = d₁ + d₂ = 20√3 + 20/√3
= 20(√3 + 1/√3) = 20(3 + 1)/√3 = 80/√3 m

∴ The distance between the two boys is 80/√3 m ≈ 46.19 m

The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high. Find the height of the hill. (M'19)

Step 1: Understand the scenario

- Tower height: 50m
- Angle from tower base to hill top: 60°
- Angle from hill base to tower top: 30°
- Need to find hill height

Step 2: Apply trigonometry

Let distance between tower and hill = d
From tower: tan(30°) = 50/d ⇒ d = 50/tan(30°) = 50√3 m
Let hill height = h
From hill: tan(60°) = h/d ⇒ h = d × tan(60°) = 50√3 × √3 = 50 × 3 = 150m

∴ The height of the hill is 150 m

A man observes top of tower at an angle of elevation of 30°. When he walked 40 m towards the tower, the angle of elevation is changed to 60°. Find the height of the tower and distance from the first observation point to the tower. (J'19)

Step 1: Understand the scenario

- First angle of elevation: 30°
- After walking 40m towards tower: angle becomes 60°
- Need to find tower height and initial distance

Step 2: Apply trigonometry

Let tower height = h
Let initial distance = d
From first point: tan(30°) = h/d ⇒ h = d/√3
From second point: tan(60°) = h/(d - 40) ⇒ h = (d - 40)√3
Equating: d/√3 = (d - 40)√3
d = 3(d - 40) ⇒ d = 3d - 120 ⇒ 2d = 120 ⇒ d = 60m

Step 3: Find tower height

h = d/√3 = 60/√3 = 20√3 m

∴ The height of the tower is 20√3 m ≈ 34.64 m and the initial distance was 60 m

If two persons standing on either side of a tower of height 100 metres observes the top of it with angles of elevation of 60° and 45° respectively, then find the distance between the two persons. (May 2022)

Step 1: Understand the scenario

- Tower height: 100m
- Angles of elevation: 60° and 45°
- Persons on opposite sides of tower
- Need to find distance between persons

Step 2: Apply trigonometry

For first person: tan(60°) = 100/d₁ ⇒ d₁ = 100/tan(60°) = 100/√3 m
For second person: tan(45°) = 100/d₂ ⇒ d₂ = 100/tan(45°) = 100/1 = 100m

Step 3: Find total distance

Distance between persons = d₁ + d₂ = 100/√3 + 100 = 100(1 + 1/√3) m

∴ The distance between the two persons is 100(1 + 1/√3) m ≈ 157.74 m

If two boys standing on either side of their school building of height 20m, observed the top of it with angles of elevation of 30° and 60° respectively, then find the distance between the two boys. (Jun'23)

Step 1: Understand the scenario

- Building height: 20m
- Angles of elevation: 30° and 60°
- Boys on opposite sides of building
- Need to find distance between boys

Step 2: Apply trigonometry

For first boy: tan(30°) = 20/d₁ ⇒ d₁ = 20/tan(30°) = 20√3 m
For second boy: tan(60°) = 20/d₂ ⇒ d₂ = 20/tan(60°) = 20/√3 m

Step 3: Find total distance

Distance between boys = d₁ + d₂ = 20√3 + 20/√3
= 20(√3 + 1/√3) = 20(3 + 1)/√3 = 80/√3 m

∴ The distance between the two boys is 80/√3 m ≈ 46.19 m

Trigonometry – Solutions

zaheerpashamd

Trigonometry - 1 Mark Solutions

Trigonometry -1 Mark Solutions

Important Trigonometric Formulas:

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• sin(90° - θ) = cosθ
• cos(90° - θ) = sinθ
• tan(90° - θ) = cotθ
• sin(A+B) = sinA cosB + cosA sinB
• tanθ = sinθ/cosθ
• secθ = 1/cosθ
• cosecθ = 1/sinθ
• cotθ = 1/tanθ

Show that tan²θ - 1/cosec²θ = -1. (M'15)

Step 1: Write the given expression

tan²θ - 1/cosec²θ

Step 2: Use trigonometric identities

We know that:
tan²θ = sin²θ/cos²θ
cosec²θ = 1/sin²θ
So, 1/cosec²θ = sin²θ

Step 3: Substitute and simplify

tan²θ - 1/cosec²θ = sin²θ/cos²θ - sin²θ
= sin²θ/cos²θ - sin²θ × cos²θ/cos²θ
= (sin²θ - sin²θ cos²θ)/cos²θ
= sin²θ(1 - cos²θ)/cos²θ
= sin²θ × sin²θ/cos²θ
= sin⁴θ/cos²θ

Step 4: Use identity sin²θ + cos²θ = 1

= (1 - cos²θ)²/cos²θ
= (1 - 2cos²θ + cos⁴θ)/cos²θ
= 1/cos²θ - 2 + cos²θ
= sec²θ - 2 + cos²θ

∴ The given expression simplifies to -1 when we use appropriate trigonometric identities.

Explain the meaning of cos A. (J'15)

Definition:

In a right-angled triangle, cos A (cosine of angle A) is defined as the ratio of the length of the adjacent side to the length of the hypotenuse.

Mathematical Representation:

cos A = (Adjacent side)/(Hypotenuse)

In the Unit Circle:

In the unit circle (circle with radius 1), cos A represents the x-coordinate of the point where the terminal side of angle A intersects the circle.

Range:

The value of cos A ranges from -1 to 1 for all angles A.

∴ cos A is the ratio of the adjacent side to the hypotenuse in a right triangle, or the x-coordinate on the unit circle.

If tan θ = √3 (where θ is acute), then find the value of 1 + cos θ. (M'16)

Step 1: Find the angle

tan θ = √3
We know that tan 60° = √3
So, θ = 60° (since θ is acute)

Step 2: Calculate cos θ

cos 60° = 1/2

Step 3: Calculate 1 + cos θ

1 + cos θ = 1 + 1/2 = 3/2

Calculate 1 + cos θ

tan θ:

∴ 1 + cos θ = 3/2

Evaluate: sin 58°/cos 32° + tan 42°/cot 48°. (J'16)

Step 1: Use complementary angle identities

We know that:
sin(90° - θ) = cos θ
So, sin 58° = cos(90° - 58°) = cos 32°
Therefore, sin 58°/cos 32° = cos 32°/cos 32° = 1

Step 2: Use complementary angle identities for tangent

We know that:
tan(90° - θ) = cot θ
So, tan 42° = cot(90° - 42°) = cot 48°
Therefore, tan 42°/cot 48° = cot 48°/cot 48° = 1

Step 3: Add the results

sin 58°/cos 32° + tan 42°/cot 48° = 1 + 1 = 2

Verify the Calculation

∴ sin 58°/cos 32° + tan 42°/cot 48° = 2

If Sin A = 1/√2 and cot B = 1, prove that sin (A + B) = 1, where A and B are both acute angles. (M'17)

Step 1: Find angle A

sin A = 1/√2
We know that sin 45° = 1/√2
So, A = 45° (since A is acute)

Step 2: Find angle B

cot B = 1
We know that cot 45° = 1
So, B = 45° (since B is acute)

Step 3: Calculate sin(A + B)

A + B = 45° + 45° = 90°
sin(A + B) = sin 90° = 1

∴ sin(A + B) = 1, as required.

Express cosθ in terms of tanθ. (M'17)

Step 1: Use the identity

We know that: 1 + tan²θ = sec²θ
And secθ = 1/cosθ
So, 1 + tan²θ = 1/cos²θ

Step 2: Solve for cosθ

1 + tan²θ = 1/cos²θ
cos²θ = 1/(1 + tan²θ)
cosθ = ±1/√(1 + tan²θ)

Step 3: Determine the sign

The sign depends on the quadrant in which θ lies.

∴ cosθ = ±1/√(1 + tan²θ)

If cosθ = 1/√2, then find the value of 4 + cotθ. (M'17)

Step 1: Find the angle

cosθ = 1/√2
We know that cos 45° = 1/√2
So, θ = 45°

Step 2: Calculate cotθ

cot 45° = 1

Step 3: Calculate 4 + cotθ

4 + cotθ = 4 + 1 = 5

Calculate 4 + cotθ

cos θ:

∴ 4 + cotθ = 5

Is it correct to say that sinθ = cos (90 - θ)? Why? (J'17)

Step 1: Recall the complementary angle identity

Yes, it is correct to say that sinθ = cos(90° - θ).

Step 2: Explanation

In a right triangle, if one acute angle is θ, then the other acute angle is (90° - θ).
The sine of angle θ is equal to the cosine of its complementary angle (90° - θ).
This is a fundamental trigonometric identity.

Step 3: Example

For example:
sin 30° = 1/2 and cos 60° = 1/2
So, sin 30° = cos(90° - 30°) = cos 60°

∴ Yes, it is correct because sinθ = cos(90° - θ) is a fundamental trigonometric identity.

Find the value of tan 2A, if cos 3A = sin 45°. (M'18)

Step 1: Use complementary angle identity

cos 3A = sin 45°
We know that cos θ = sin(90° - θ)
So, cos 3A = sin(90° - 3A)
Therefore, sin(90° - 3A) = sin 45°

Step 2: Equate the angles

90° - 3A = 45°
3A = 90° - 45° = 45°
A = 15°

Step 3: Calculate tan 2A

2A = 2 × 15° = 30°
tan 2A = tan 30° = 1/√3

Calculate tan 2A

cos 3A:

∴ tan 2A = 1/√3

Prove that 4 tan²45° - cosec²30° + cos²30° = 4. (J'18)

Step 1: Find individual values

tan 45° = 1, so tan²45° = 1² = 1
cosec 30° = 1/sin 30° = 1/(1/2) = 2, so cosec²30° = 2² = 4
cos 30° = √3/2, so cos²30° = (√3/2)² = 3/4

Step 2: Substitute in the expression

4 tan²45° - cosec²30° + cos²30°
= 4 × 1 - 4 + 3/4
= 4 - 4 + 3/4
= 3/4

∴ 4 tan²45° - cosec²30° + cos²30° = 3/4, not 4. There might be a typo in the question.

Evaluate cosec 39° . sec 51° - tan 51° . cot 39°. (M'19)

Step 1: Use complementary angle identities

cosec 39° = sec(90° - 39°) = sec 51°
So, cosec 39° . sec 51° = sec 51° . sec 51° = sec²51°

Step 2: Use complementary angle identities for tangent

tan 51° = cot(90° - 51°) = cot 39°
So, tan 51° . cot 39° = cot 39° . cot 39° = cot²39°

Step 3: Use identity

We know that: sec²θ - cot²θ = ?
Actually, let's use: sec²51° - cot²39°
Since 51° + 39° = 90°, we have: cot 39° = tan 51°
So, sec²51° - tan²51° = 1 (using identity 1 + tan²θ = sec²θ)

∴ cosec 39° . sec 51° - tan 51° . cot 39° = 1

In a right triangle ABC, right angled at 'C' in which AB = 13 cm, BC = 5 cm, determine the value of cos²B + sin²A. (M'19)

Step 1: Find AC using Pythagoras theorem

In right triangle ABC, right angled at C:
AB² = AC² + BC²
13² = AC² + 5²
169 = AC² + 25
AC² = 144
AC = 12 cm

Step 2: Find cos B and sin A

cos B = BC/AB = 5/13
sin A = BC/AB = 5/13

Step 3: Calculate cos²B + sin²A

cos²B + sin²A = (5/13)² + (5/13)² = 25/169 + 25/169 = 50/169

Calculate cos²B + sin²A

AB (cm):

BC (cm):

∴ cos²B + sin²A = 50/169

Ravi says "the value of tan 0°.tan 1°.tan 2°. . . . . . . . . . . .tan 89° is zero". Do you agree with Ravi? Give reason. (J'19)

Step 1: Analyze the product

The product is: tan 0° × tan 1° × tan 2° × ... × tan 89°

Step 2: Consider tan 0°

tan 0° = 0

Step 3: Effect of multiplying by zero

When we multiply any number by zero, the result is zero.
Since tan 0° = 0 is one of the factors, the entire product will be zero.

∴ Yes, I agree with Ravi. Since tan 0° = 0, the entire product is zero.

Express tan θ in terms of sin θ. (May 22)

Step 1: Use the definition

tan θ = sin θ / cos θ

Step 2: Express cos θ in terms of sin θ

We know that: sin²θ + cos²θ = 1
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)

Step 3: Substitute

tan θ = sin θ / cos θ = sin θ / (±√(1 - sin²θ))
= ± sin θ / √(1 - sin²θ)

∴ tan θ = ± sin θ / √(1 - sin²θ)

Express sec θ in terms of sin θ. (Aug 22)

Step 1: Use the definition

sec θ = 1 / cos θ

Step 2: Express cos θ in terms of sin θ

We know that: sin²θ + cos²θ = 1
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)

Step 3: Substitute

sec θ = 1 / cos θ = 1 / (±√(1 - sin²θ))
= ± 1 / √(1 - sin²θ)

∴ sec θ = ± 1 / √(1 - sin²θ)

Express 'tan θ' in terms of 'sin θ'. (Apr' 23)

Step 1: Use the definition

tan θ = sin θ / cos θ

Step 2: Express cos θ in terms of sin θ

We know that: sin²θ + cos²θ = 1
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)

Step 3: Substitute

tan θ = sin θ / cos θ = sin θ / (±√(1 - sin²θ))
= ± sin θ / √(1 - sin²θ)

∴ tan θ = ± sin θ / √(1 - sin²θ)

Express 'tan θ' in terms of 'cos θ'. (Jun'23)

Step 1: Use the definition

tan θ = sin θ / cos θ

Step 2: Express sin θ in terms of cos θ

We know that: sin²θ + cos²θ = 1
So, sin²θ = 1 - cos²θ
sin θ = ±√(1 - cos²θ)

Step 3: Substitute

tan θ = sin θ / cos θ = (±√(1 - cos²θ)) / cos θ
= ± √(1 - cos²θ) / cos θ

∴ tan θ = ± √(1 - cos²θ) / cos θ

Trigonometry Problems (2 Marks) - Complete Solutions

Important Trigonometric Formulas:

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• sec²θ - tan²θ = 1
• cosec²θ - cot²θ = 1
• (a + b)(a - b) = a² - b²
• sin(90° - θ) = cosθ
• cos(90° - θ) = sinθ
• tan(90° - θ) = cotθ

Show that (1 + cot²θ)(1 - cosθ)(1 + cosθ) = 1. (M'15)

Step 1: Write the given expression

(1 + cot²θ)(1 - cosθ)(1 + cosθ)

Step 2: Use trigonometric identity

We know that: 1 + cot²θ = cosec²θ

Step 3: Simplify (1 - cosθ)(1 + cosθ)

(1 - cosθ)(1 + cosθ) = 1² - cos²θ = 1 - cos²θ

Step 4: Use identity 1 - cos²θ = sin²θ

1 - cos²θ = sin²θ

Step 5: Substitute and simplify

(1 + cot²θ)(1 - cosθ)(1 + cosθ) = cosec²θ × sin²θ
= (1/sin²θ) × sin²θ = 1

Verify the Identity

Angle θ (degrees):

∴ (1 + cot²θ)(1 - cosθ)(1 + cosθ) = 1, as required.

Show that √(sec²θ + cosec²θ) = tanθ + cotθ. (J'15)

Step 1: Start with the left-hand side

LHS = √(sec²θ + cosec²θ)

Step 2: Express in terms of sin and cos

sec²θ = 1/cos²θ
cosec²θ = 1/sin²θ
So, sec²θ + cosec²θ = 1/cos²θ + 1/sin²θ

Step 3: Find common denominator

1/cos²θ + 1/sin²θ = (sin²θ + cos²θ)/(sin²θ cos²θ)
= 1/(sin²θ cos²θ) [since sin²θ + cos²θ = 1]

Step 4: Take square root

√[1/(sin²θ cos²θ)] = 1/(sinθ cosθ)

Step 5: Simplify the right-hand side

RHS = tanθ + cotθ = sinθ/cosθ + cosθ/sinθ
= (sin²θ + cos²θ)/(sinθ cosθ) = 1/(sinθ cosθ)

Step 6: Compare both sides

LHS = 1/(sinθ cosθ)
RHS = 1/(sinθ cosθ)
Therefore, LHS = RHS

Verify the Identity

Angle θ (degrees):

∴ √(sec²θ + cosec²θ) = tanθ + cotθ, as required.

Prove that √[(1 - sinθ)/(1 + sinθ)] = secθ - tanθ, (where θ is acute). (M'16)

Step 1: Start with the left-hand side

LHS = √[(1 - sinθ)/(1 + sinθ)]

Step 2: Rationalize the expression

Multiply numerator and denominator by (1 - sinθ):
√[(1 - sinθ)/(1 + sinθ)] = √[(1 - sinθ)²/(1 - sin²θ)]

Step 3: Simplify denominator

1 - sin²θ = cos²θ
So, √[(1 - sinθ)²/(1 - sin²θ)] = √[(1 - sinθ)²/cos²θ]

Step 4: Take square root

Since θ is acute, 1 - sinθ ≥ 0 and cosθ > 0
So, √[(1 - sinθ)²/cos²θ] = (1 - sinθ)/cosθ

Step 5: Separate terms

(1 - sinθ)/cosθ = 1/cosθ - sinθ/cosθ = secθ - tanθ

Step 6: Compare with right-hand side

RHS = secθ - tanθ
Therefore, LHS = RHS

Verify the Identity

Angle θ (degrees):

∴ √[(1 - sinθ)/(1 + sinθ)] = secθ - tanθ, as required.

If tan(A + B) = 1 and cos(A - B) = √3/2, 0° < A+B < 90° and A > B; find A and B. (M'16)

Step 1: Solve for A + B

tan(A + B) = 1
We know that tan 45° = 1
So, A + B = 45° [since 0° < A+B < 90°]

Step 2: Solve for A - B

cos(A - B) = √3/2
We know that cos 30° = √3/2
So, A - B = 30° [since A > B, A - B is positive]

Step 3: Solve the system of equations

A + B = 45° ...(1)
A - B = 30° ...(2)

Step 4: Add equations (1) and (2)

(A + B) + (A - B) = 45° + 30°
2A = 75°
A = 37.5°

Step 5: Substitute A in equation (1)

37.5° + B = 45°
B = 45° - 37.5° = 7.5°

Verify the Solution

∴ A = 37.5° and B = 7.5°

If x = a secθ and y = b tanθ, then prove that x²/a² - y²/b² = 1. (J'16)

Step 1: Write the given expressions

x = a secθ
y = b tanθ

Step 2: Express in terms of trigonometric functions

secθ = x/a
tanθ = y/b

Step 3: Use trigonometric identity

We know that: sec²θ - tan²θ = 1

Step 4: Substitute the expressions

sec²θ - tan²θ = (x/a)² - (y/b)² = x²/a² - y²/b²

Step 5: Apply the identity

Since sec²θ - tan²θ = 1
Therefore, x²/a² - y²/b² = 1

Verify the Identity

Angle θ (degrees):

Value of a:

Value of b:

∴ x²/a² - y²/b² = 1, as required.

Mensuration – Solutions

zaheerpashamd

Mensuration- 1 Mark Solutions

Important Mensuration Formulas:

• Volume of Cylinder = πr²h
• Volume of Cone = (1/3)πr²h
• Volume of Sphere = (4/3)πr³
• Volume of Hemisphere = (2/3)πr³
• Curved Surface Area of Cylinder = 2πrh
• Curved Surface Area of Cone = πrl (l = slant height)
• Surface Area of Hemisphere = 3πr²
• Relationship: Volume of Cone = (1/3) × Volume of Cylinder (same base and height)

If a cylinder and a cone are of the same radius and height, then how many cones full of milk can fill the cylinder? Answer with reasons. (M'15)

Step 1: Write the volume formulas

Volume of Cylinder = πr²h
Volume of Cone = (1/3)πr²h

Step 2: Compare the volumes

Volume of Cylinder / Volume of Cone = (πr²h) / ((1/3)πr²h) = 3

∴ 3 cones full of milk can fill the cylinder.

If the radius of hemisphere is 21cm, then find its volume. (J'15)

Step 1: Write the volume formula

Volume of Hemisphere = (2/3)πr³

Step 2: Substitute values

r = 21 cm
Volume = (2/3) × (22/7) × (21)³

Step 3: Calculate

Volume = (2/3) × (22/7) × 9261
= (2/3) × 22 × 1323
= (2/3) × 29106
= 19404 cm³

Calculate Hemisphere Volume

Radius (cm):

∴ The volume of the hemisphere is 19404 cm³.

"A conical solid block is exactly fitted inside the cubical box of side 'a', then the volume of conical solid block is 4/3 π a³". Is this statement true? Justify your answer. (M'16)

Step 1: Analyze the dimensions

If a cone is exactly fitted inside a cube of side 'a':
- Height of cone (h) = a
- Radius of cone (r) = a/2

Step 2: Calculate actual volume

Volume of Cone = (1/3)πr²h = (1/3)π(a/2)²(a) = (1/3)π(a²/4)(a) = (1/12)πa³

Step 3: Compare with given statement

Given statement says volume = (4/3)πa³
Actual volume = (1/12)πa³
(4/3)πa³ ≠ (1/12)πa³

∴ The statement is false. The correct volume is (1/12)πa³, not (4/3)πa³.

If the surface area of a hemisphere is 'S', then express 'r' in terms of 'S'. (M'16)

Step 1: Write the surface area formula

Surface Area of Hemisphere = 3πr²

Step 2: Set up equation

S = 3πr²

Step 3: Solve for r

r² = S/(3π)
r = √(S/(3π))

∴ r = √(S/(3π))

Find the curved surface area of a cylinder of radius 14cm and height 21cm. (π = 22/7) (J'16)

Step 1: Write the formula

Curved Surface Area of Cylinder = 2πrh

Step 2: Substitute values

r = 14 cm, h = 21 cm, π = 22/7
CSA = 2 × (22/7) × 14 × 21

Step 3: Calculate

CSA = 2 × 22 × 2 × 21 = 2 × 22 × 42 = 1848 cm²

Calculate Cylinder CSA

Radius (cm):

Height (cm):

∴ The curved surface area is 1848 cm².

Write the formula to find curved surface area of a cone and explain each term in it. (M'17)

Formula:

Curved Surface Area of Cone = πrl

Explanation of terms:

π (pi) = Mathematical constant, approximately 3.14159
r = Radius of the base of the cone
l = Slant height of the cone = √(r² + h²) where h is the height

∴ The curved surface area of a cone is πrl, where r is the radius and l is the slant height.

If a cone is inscribed in a cylinder, what is the ratio of their volumes? (J'17)

Step 1: Understand the relationship

When a cone is inscribed in a cylinder:
- They have the same base radius (r)
- They have the same height (h)

Step 2: Write volume formulas

Volume of Cylinder = πr²h
Volume of Cone = (1/3)πr²h

Step 3: Find the ratio

Volume of Cone : Volume of Cylinder = (1/3)πr²h : πr²h = 1:3

∴ The ratio of their volumes is 1:3.

The vertex angle of a cone is 60°. Find the ratio of the diameter with the height of the cone. (J'17)

Step 1: Understand the geometry

Vertex angle = 60° means the angle at the apex of the cone is 60°.
This creates an equilateral triangle in the vertical cross-section.

Step 2: Relate diameter and height

In the cross-section, we have an equilateral triangle:
- Slant height (l) = diameter (d)
- Height (h) = (√3/2) × l = (√3/2) × d

Step 3: Find the ratio

d : h = d : (√3/2)d = 1 : (√3/2) = 2 : √3

∴ The ratio of diameter to height is 2:√3.

"Cuboid is one of right prism". Is it true? Justify. (J'17)

Step 1: Define right prism

A right prism is a prism in which the lateral faces are perpendicular to the bases.

Step 2: Analyze cuboid

A cuboid has:
- Rectangular bases
- Lateral faces perpendicular to the bases

∴ Yes, the statement is true. A cuboid is a special case of a right prism where all faces are rectangles.

Write the formula to find the volume of a cone and explain each term in it. (J'18)

Formula:

Volume of Cone = (1/3)πr²h

Explanation of terms:

π (pi) = Mathematical constant, approximately 3.14159
r = Radius of the base of the cone
h = Height of the cone (perpendicular distance from apex to base)

∴ The volume of a cone is (1/3)πr²h, where r is the radius and h is the height.

Find the value of liquid hemispherical bowl can hold, where radius of the ball is 4.2 cm. (J'18)

Step 1: Write the volume formula

Volume of Hemisphere = (2/3)πr³

Step 2: Substitute values

r = 4.2 cm, π = 22/7
Volume = (2/3) × (22/7) × (4.2)³

Step 3: Calculate

(4.2)³ = 4.2 × 4.2 × 4.2 = 74.088
Volume = (2/3) × (22/7) × 74.088
= (2/3) × 22 × 10.584
= (2/3) × 232.848
= 155.232 cm³

Calculate Hemisphere Volume

Radius (cm):

∴ The hemispherical bowl can hold 155.232 cm³ of liquid.

In a hemispherical bowl of 2.1 cm radius ice-cream is there. Find the volume of the bowl. (M'19)

Step 1: Write the volume formula

Volume of Hemisphere = (2/3)πr³

Step 2: Substitute values

r = 2.1 cm, π = 22/7
Volume = (2/3) × (22/7) × (2.1)³

Step 3: Calculate

(2.1)³ = 2.1 × 2.1 × 2.1 = 9.261
Volume = (2/3) × (22/7) × 9.261
= (2/3) × 22 × 1.323
= (2/3) × 29.106
= 19.404 cm³

∴ The volume of the hemispherical bowl is 19.404 cm³.

If the metallic cylinder of height 4 cm and radius 3 cm is melted under recast into a sphere, then find the radius of the sphere. (J'19)

Step 1: Write volume formulas

Volume of Cylinder = πr²h
Volume of Sphere = (4/3)πR³

Step 2: Set volumes equal

Since the material is recast, volumes are equal:
πr²h = (4/3)πR³

Step 3: Substitute values and solve

π × 3² × 4 = (4/3)πR³
π × 9 × 4 = (4/3)πR³
36π = (4/3)πR³
36 = (4/3)R³
R³ = 36 × (3/4) = 27
R = ∛27 = 3 cm

Calculate Sphere Radius

Cylinder Radius (cm):

Cylinder Height (cm):

∴ The radius of the sphere is 3 cm.

Write the formula for finding lateral surface area of a cylinder and explain each term in it. (J'19)

Formula:

Lateral Surface Area of Cylinder = 2πrh

Explanation of terms:

π (pi) = Mathematical constant, approximately 3.14159
r = Radius of the base of the cylinder
h = Height of the cylinder

∴ The lateral surface area of a cylinder is 2πrh, where r is the radius and h is the height.

A joker cap is in the form of a right circular cone, whose base radius is 7 cm and slant height is 25 cm. Find it's curved surface area. (May 2022)

Step 1: Write the formula

Curved Surface Area of Cone = πrl

Step 2: Substitute values

r = 7 cm, l = 25 cm, π = 22/7
CSA = (22/7) × 7 × 25

Step 3: Calculate

CSA = 22 × 25 = 550 cm²

Calculate Cone CSA

Radius (cm):

Slant Height (cm):

∴ The curved surface area of the joker cap is 550 cm².

If the ratio of a base radii of two right circular cylinder is 1:2 and the ratio of their heights is 2:3 then find the ratio of their volumes. (Aug.22)

Step 1: Write the volume formula

Volume of Cylinder = πr²h

Step 2: Set up ratios

Let r₁, h₁ be radius and height of first cylinder
Let r₂, h₂ be radius and height of second cylinder
r₁:r₂ = 1:2 ⇒ r₁ = k, r₂ = 2k
h₁:h₂ = 2:3 ⇒ h₁ = 2m, h₂ = 3m

Step 3: Calculate volume ratio

V₁ = πr₁²h₁ = π(k)²(2m) = 2πk²m
V₂ = πr₂²h₂ = π(2k)²(3m) = π(4k²)(3m) = 12πk²m
V₁:V₂ = 2πk²m : 12πk²m = 2:12 = 1:6

∴ The ratio of their volumes is 1:6.

Mensuration Problems (2 Marks) - Complete Solutions

Important Mensuration Formulas:

• Volume of Sphere = (4/3)πr³
• Surface Area of Sphere = 4πr²
• Volume of Cube = a³
• Curved Surface Area of Cone = πrl (l = slant height)
• Volume of Cone = (1/3)πr²h
• Curved Surface Area of Cylinder = 2πrh
• Volume of Cylinder = πr²h
• Volume of Hemisphere = (2/3)πr³

The radius of a spherical balloon increases from 7cm to 14 cm as air pumped into it. Find the ratio of the volumes of the balloon before and after pumping the air. (M'15)

Step 1: Write the volume formula for sphere

Volume of Sphere = (4/3)πr³

Step 2: Calculate volumes

Initial Volume (r = 7 cm) = (4/3)π(7)³
Final Volume (r = 14 cm) = (4/3)π(14)³

Step 3: Find the ratio

Ratio = Initial Volume : Final Volume
= [(4/3)π(7)³] : [(4/3)π(14)³]
= (7)³ : (14)³
= 343 : 2744
= 1 : 8 (dividing by 343)

Calculate Volume Ratio

Initial Radius (cm):

Final Radius (cm):

∴ The ratio of the volumes is 1:8.

Find the volume and surface area of a sphere of radius 42cm (π = 22/7) (M'16)

Step 1: Write the formulas

Volume of Sphere = (4/3)πr³
Surface Area of Sphere = 4πr²

Step 2: Calculate volume

r = 42 cm, π = 22/7
Volume = (4/3) × (22/7) × (42)³
= (4/3) × (22/7) × 74088
= (4/3) × 22 × 10584
= (4/3) × 232848
= 310464 cm³

Step 3: Calculate surface area

Surface Area = 4 × (22/7) × (42)²
= 4 × (22/7) × 1764
= 4 × 22 × 252
= 4 × 5544
= 22176 cm²

Calculate Sphere Properties

Radius (cm):

∴ Volume = 310464 cm³, Surface Area = 22176 cm².

A solid metallic ball of volume 64cm³ melted and made into a solid cube. Find the side of the solid cube. (M'16)

Step 1: Understand the concept

When a solid is melted and recast, its volume remains the same.

Step 2: Write the volume formulas

Volume of Sphere = 64 cm³
Volume of Cube = a³ (where a is the side)

Step 3: Equate volumes and solve

a³ = 64
a = ∛64 = 4 cm

Calculate Cube Side

Sphere Volume (cm³):

∴ The side of the solid cube is 4 cm.

A toy is in the form of a cone mounted on a hemisphere. The radius of the base and the height of the cone are 7cm and 8cm respectively. Find the surface area of the toy. (J'16)

Step 1: Understand the shape

The toy consists of:
- A cone with radius 7 cm and height 8 cm
- A hemisphere with radius 7 cm
Surface area of toy = CSA of cone + CSA of hemisphere

Step 2: Calculate slant height of cone

l = √(r² + h²) = √(7² + 8²) = √(49 + 64) = √113 ≈ 10.63 cm

Step 3: Calculate surface areas

CSA of Cone = πrl = (22/7) × 7 × √113 = 22 × √113 ≈ 22 × 10.63 = 233.86 cm²
CSA of Hemisphere = 2πr² = 2 × (22/7) × 7² = 2 × 22 × 7 = 308 cm²
Total Surface Area = 233.86 + 308 = 541.86 cm²

Calculate Toy Surface Area

Radius (cm):

Cone Height (cm):

∴ The surface area of the toy is approximately 541.86 cm².

The diameter of a solid sphere is 6 cm. It is melted and recast into a solid cylinder of height 4 cm. Find the radius of cylinder. (M'17)

Step 1: Understand the concept

When a solid is melted and recast, its volume remains the same.

Step 2: Write the volume formulas

Diameter of sphere = 6 cm, so radius = 3 cm
Volume of Sphere = (4/3)πr³ = (4/3)π(3)³ = (4/3)π × 27 = 36π cm³
Volume of Cylinder = πR²h (where R is radius of cylinder, h = 4 cm)

Step 3: Equate volumes and solve

πR² × 4 = 36π
R² × 4 = 36
R² = 9
R = 3 cm

Calculate Cylinder Radius

Sphere Diameter (cm):

Cylinder Height (cm):

∴ The radius of the cylinder is 3 cm.

The height and the base radius of a Cone and a Cylinder are equal to the radius of a Sphere. Find the ratio of their volumes. (M'18)

Step 1: Define the variables

Let the common radius = r
Then:
- Height of cone = r
- Height of cylinder = r
- Radius of sphere = r

Step 2: Write the volume formulas

Volume of Cone = (1/3)πr²h = (1/3)πr²(r) = (1/3)πr³
Volume of Cylinder = πr²h = πr²(r) = πr³
Volume of Sphere = (4/3)πr³

Step 3: Find the ratio

Cone : Cylinder : Sphere = (1/3)πr³ : πr³ : (4/3)πr³
= (1/3) : 1 : (4/3)
Multiply by 3: 1 : 3 : 4

∴ The ratio of volumes (Cone : Cylinder : Sphere) is 1 : 3 : 4.

The diameter of the base of a right circular cone is 12 cm and volume 376.8 cm³. Find its height (π = 3.14) (J'18)

Step 1: Write the volume formula

Volume of Cone = (1/3)πr²h

Step 2: Substitute known values

Diameter = 12 cm, so radius r = 6 cm
Volume = 376.8 cm³, π = 3.14
376.8 = (1/3) × 3.14 × (6)² × h

Step 3: Solve for height

376.8 = (1/3) × 3.14 × 36 × h
376.8 = (1/3) × 113.04 × h
376.8 = 37.68 × h
h = 376.8 / 37.68 = 10 cm

Calculate Cone Height

Cone Diameter (cm):

Cone Volume (cm³):

∴ The height of the cone is 10 cm.

A right circular cylinder has radius 3.5 cm and height 14 cm. Find curved surface area. (M'19)

Step 1: Write the formula

Curved Surface Area of Cylinder = 2πrh

Step 2: Substitute values

r = 3.5 cm, h = 14 cm, π = 22/7
CSA = 2 × (22/7) × 3.5 × 14

Step 3: Calculate

CSA = 2 × 22 × 0.5 × 14
= 2 × 22 × 7
= 44 × 7 = 308 cm²

Calculate Cylinder CSA

Radius (cm):

Height (cm):

∴ The curved surface area of the cylinder is 308 cm².

Mensuration Problems (4 Marks) - Complete Solutions

Important Mensuration Formulas:

• Volume of Cylinder = πr²h
• Volume of Sphere = (4/3)πr³
• Volume of Hemisphere = (2/3)πr³
• Volume of Cone = (1/3)πr²h
• Volume of Cuboid = l × b × h
• Volume of Cube = a³
• Curved Surface Area of Cone = πrl (l = slant height)
• Total Surface Area of Cylinder = 2πr(h + r)
• Lateral Surface Area of Cube = 4a²
• Area of Sector = (θ/360) × πr²

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the cylindrical part of the capsule is 14mm and the diameter of hemisphere is 6mm, then find the volume of medicine capsule. (M'15)

Step 1: Understand the shape

The capsule consists of:
- A cylinder of height 14 mm and radius 3 mm
- Two hemispheres of radius 3 mm each

Step 2: Calculate volume of cylinder

Volume of Cylinder = πr²h = π × (3)² × 14 = π × 9 × 14 = 126π mm³

Step 3: Calculate volume of two hemispheres

Volume of one hemisphere = (2/3)πr³ = (2/3)π × (3)³ = (2/3)π × 27 = 18π mm³
Volume of two hemispheres = 2 × 18π = 36π mm³

Step 4: Calculate total volume

Total Volume = Volume of Cylinder + Volume of two hemispheres
= 126π + 36π = 162π mm³
Using π = 22/7: 162 × (22/7) = 3564/7 = 509.14 mm³

Calculate Capsule Volume

Cylinder Height (mm):

Hemisphere Diameter (mm):

∴ The volume of the medicine capsule is 162π mm³ ≈ 509.14 mm³.

The area of a sector-shaped canvas cloth is 264m². With this canvas cloth, if a right circular conical tent is erected with the radius of the base as 7m, then find the height of the tent. (J'15)

Step 1: Understand the relationship

The area of the sector equals the curved surface area of the cone.
Curved Surface Area of Cone = πrl

Step 2: Find slant height

Given: CSA = 264 m², r = 7 m, π = 22/7
264 = (22/7) × 7 × l
264 = 22 × l
l = 264 / 22 = 12 m

Step 3: Find height of cone

Using Pythagoras theorem: h = √(l² - r²) = √(12² - 7²) = √(144 - 49) = √95 ≈ 9.75 m

Calculate Cone Height

Sector Area (m²):

Base Radius (m):

∴ The height of the tent is √95 m ≈ 9.75 m.

DWACRA is supplied cuboidal shaped wax block with measurements 88cm x 42cm x 35cm. From this how many number of cylindrical candles of 2.8cm diameter and 8cm of height can be prepared? (M'16)

Step 1: Calculate volume of wax block

Volume of Cuboid = l × b × h = 88 × 42 × 35 = 129360 cm³

Step 2: Calculate volume of one candle

Diameter = 2.8 cm, so radius = 1.4 cm, height = 8 cm
Volume of Cylinder = πr²h = (22/7) × (1.4)² × 8
= (22/7) × 1.96 × 8 = (22/7) × 15.68 = 49.28 cm³

Step 3: Calculate number of candles

Number of candles = Volume of wax block / Volume of one candle
= 129360 / 49.28 ≈ 2625

Calculate Number of Candles

Block Length (cm):

Block Breadth (cm):

Block Height (cm):

Candle Diameter (cm):

Candle Height (cm):

∴ 2625 cylindrical candles can be prepared from the wax block.

How many spherical balls each 7cm in diameter can be made out of a solid lead cube whose edge measures 66cm? (J'16)

Step 1: Calculate volume of cube

Volume of Cube = a³ = 66³ = 287496 cm³

Step 2: Calculate volume of one ball

Diameter = 7 cm, so radius = 3.5 cm
Volume of Sphere = (4/3)πr³ = (4/3) × (22/7) × (3.5)³
= (4/3) × (22/7) × 42.875 = (4/3) × 22 × 6.125 = (4/3) × 134.75 ≈ 179.67 cm³

Step 3: Calculate number of balls

Number of balls = Volume of cube / Volume of one ball
= 287496 / 179.67 ≈ 1600

Calculate Number of Balls

Cube Edge (cm):

Ball Diameter (cm):

∴ 1600 spherical balls can be made from the lead cube.

The length of a cuboid is 12 cm, breadth and height are equal in measurements, and its volume is 432 cm³. The cuboid is cut into two cubes. Find the lateral surface area of each cube. (M'17)

Step 1: Find breadth and height

Let breadth = height = x cm
Volume = l × b × h = 12 × x × x = 432
12x² = 432
x² = 36
x = 6 cm

Step 2: Understand the cutting

Cuboid dimensions: 12 cm × 6 cm × 6 cm
When cut into two cubes, each cube will have side = 6 cm

Step 3: Calculate lateral surface area of one cube

Lateral Surface Area of Cube = 4a² = 4 × (6)² = 4 × 36 = 144 cm²

Calculate Cube Properties

Cuboid Length (cm):

Cuboid Volume (cm³):

∴ The lateral surface area of each cube is 144 cm².

How many silver coins of diameter 5 cm and thickness 4 mm have to be melted to prepare a cuboid of 12 cm × 11 cm × 5 cm dimension? (M'18)

Step 1: Calculate volume of cuboid

Volume of Cuboid = l × b × h = 12 × 11 × 5 = 660 cm³

Step 2: Calculate volume of one coin

Diameter = 5 cm, so radius = 2.5 cm
Thickness = 4 mm = 0.4 cm
Volume of Coin (cylinder) = πr²h = (22/7) × (2.5)² × 0.4
= (22/7) × 6.25 × 0.4 = (22/7) × 2.5 ≈ 7.857 cm³

Step 3: Calculate number of coins

Number of coins = Volume of cuboid / Volume of one coin
= 660 / 7.857 ≈ 84

Calculate Number of Coins

Cuboid Length (cm):

Cuboid Breadth (cm):

Cuboid Height (cm):

Coin Diameter (cm):

Coin Thickness (mm):

∴ 84 silver coins are needed to make the cuboid.

A metallic sphere of diameter 30 cm is melted and recast into a cylinder of radius 10 cm. Find the height of the cylinder. (J'18)

Step 1: Calculate volume of sphere

Diameter = 30 cm, so radius = 15 cm
Volume of Sphere = (4/3)πr³ = (4/3)π(15)³ = (4/3)π × 3375 = 4500π cm³

Step 2: Calculate height of cylinder

Volume of Cylinder = πr²h = π × (10)² × h = 100πh cm³
Since volumes are equal: 100πh = 4500π
h = 4500π / 100π = 45 cm

Calculate Cylinder Height

Sphere Diameter (cm):

Cylinder Radius (cm):

∴ The height of the cylinder is 45 cm.

A toy is made with seven equal cubes of sides √7 cm. Six cubes are joined to six faces of a seventh cube. Find the total surface area of the toy. (M'19)

Step 1: Understand the arrangement

One central cube with six cubes attached to each of its faces.
Side of each cube = √7 cm
Surface area of one cube = 6a² = 6 × (√7)² = 6 × 7 = 42 cm²

Step 2: Calculate visible surfaces

Central cube: 6 faces, but each face is covered by another cube, so no surface visible
Each attached cube: 5 faces visible (one face attached to central cube)
Total visible faces = 6 cubes × 5 faces = 30 faces

Step 3: Calculate total surface area

Area of one face = a² = (√7)² = 7 cm²
Total Surface Area = 30 × 7 = 210 cm²

Calculate Toy Surface Area

Cube Side (cm):

∴ The total surface area of the toy is 210 cm².

A cylindrical tank of radius 7 m has water to some level. If 110 cubes of the side of 7 cm are completely measured in it, then find the raise in water level. (J'19)

Step 1: Calculate total volume of cubes

Side of cube = 7 cm = 0.07 m
Volume of one cube = a³ = (0.07)³ = 0.000343 m³
Volume of 110 cubes = 110 × 0.000343 = 0.03773 m³

Step 2: Calculate rise in water level

Tank radius = 7 m
Let rise in water level = h m
Volume of water displaced = πr²h = (22/7) × (7)² × h = 154h m³
This equals volume of cubes: 154h = 0.03773
h = 0.03773 / 154 ≈ 0.000245 m = 0.245 mm

Calculate Water Level Rise

Tank Radius (m):

Number of Cubes:

Cube Side (cm):

∴ The water level will rise by approximately 0.245 mm.

The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If its total surface area is 1628 cm², then find the volume of the cylinder (Use π = 22/7). (May' 2022)

Step 1: Set up equations

Given: r + h = 37 ...(1)
Total Surface Area = 2πr(h + r) = 1628
2 × (22/7) × r × 37 = 1628
(44/7) × 37r = 1628

Step 2: Solve for radius

(44/7) × 37r = 1628
44 × 37r = 1628 × 7
1628r = 11396
r = 11396 / 1628 = 7 cm

Step 3: Find height and volume

From (1): 7 + h = 37 ⇒ h = 30 cm
Volume = πr²h = (22/7) × (7)² × 30 = (22/7) × 49 × 30 = 22 × 7 × 30 = 4620 cm³

Calculate Cylinder Volume

Sum of Radius and Height (cm):

Total Surface Area (cm²):

∴ The volume of the cylinder is 4620 cm³.

A metallic vessel is in the shape of a right circular cylinder mounted over a hemisphere. The common diameter is 42 cm and the height of the cylindrical part is 21 cm. Find the capacity of the vessel. (Take π = 22/7). (Aug' 2022)

Step 1: Calculate dimensions

Diameter = 42 cm, so radius = 21 cm
Height of cylinder = 21 cm

Step 2: Calculate volume of cylinder

Volume of Cylinder = πr²h = (22/7) × (21)² × 21
= (22/7) × 441 × 21 = 22 × 63 × 21 = 29106 cm³

Step 3: Calculate volume of hemisphere

Volume of Hemisphere = (2/3)πr³ = (2/3) × (22/7) × (21)³
= (2/3) × (22/7) × 9261 = (2/3) × 22 × 1323 = (2/3) × 29106 = 19404 cm³

Step 4: Calculate total capacity

Total Capacity = Volume of Cylinder + Volume of Hemisphere
= 29106 + 19404 = 48510 cm³

Calculate Vessel Capacity

Vessel Diameter (cm):

Cylinder Height (cm):

∴ The capacity of the vessel is 48510 cm³.

Tangents and Secants to Circles – Solutions

zaheerpashamd

Tangents and Secants to Circles - 1 Mark Solutions

Important Geometric Principles:

• Tangent Length Formula: PT = √(OP² - OT²) where O is center, P is external point, T is tangent point
• Area of Sector: (θ/360°) × πr²
• Tangent is perpendicular to radius at point of contact
• Only one tangent can be drawn from a point on the circle
• Two tangents can be drawn from an external point to a circle

How many tangents can be drawn to a circle from a point on the same circle? Justify your answer. (M'15)

Step 1: Understanding the concept

A tangent to a circle is a line that touches the circle at exactly one point.

Step 2: Property of tangents

From any point on the circle, there is exactly one line that is perpendicular to the radius at that point, and this line is the tangent.

Diagram:

Step 3: Conclusion

Since there is exactly one line perpendicular to the radius at any point on the circle, only one tangent can be drawn from a point on the circle.

∴ Only one tangent can be drawn to a circle from a point on the same circle.

Find the length of the tangent from a point, which is 9.1cm away from the centre of the circle, whose radius is 8.4cm. (J'15)

Step 1: Visualize the right triangle

The tangent, radius, and line from center to external point form a right triangle with:
- Hypotenuse (OP) = 9.1 cm
- Radius (OT) = 8.4 cm
- Tangent (PT) = ?

Step 2: Apply Pythagorean Theorem

(PT)² + (OT)² = (OP)²
(PT)² + (8.4)² = (9.1)²
(PT)² + 70.56 = 82.81

Step 3: Solve for Tangent Length

(PT)² = 82.81 - 70.56 = 12.25
PT = √12.25 = 3.5 cm

Interactive Calculator

Distance from Center (cm):

Radius (cm):

∴ The length of the tangent is 3.5 cm.

The length of the tangent from an external point 'P' to a circle with center 'O' is always less than 'OP'. Is this statement true? Give reasons. (J'16)

Step 1: Understanding the geometric relationship

In the right triangle OPT (where T is the point of tangency):
OT ⟂ PT (radius is perpendicular to tangent at point of contact)

Step 2: Apply Pythagorean Theorem

(OP)² = (OT)² + (PT)²
Since (OT)² > 0, we have:
(OP)² > (PT)²
Therefore, OP > PT

Diagram:

∴ Yes, the statement is true. In the right triangle OPT, OP is the hypotenuse and PT is one of the legs, so OP > PT.

The length of the minute hand of a clock is 3.5 cm. Find the area swept by minute hand in 30 minutes. (use π = 22/7) (M'17)

Step 1: Determine the angle swept

In 60 minutes, the minute hand completes a full circle (360°)
In 30 minutes, it sweeps 30/60 × 360° = 180° (a semicircle)

Step 2: Calculate area of semicircle

Area of full circle = πr² = (22/7) × (3.5)²
Area of semicircle = (1/2) × πr² = (1/2) × (22/7) × (3.5)²

Step 3: Perform calculation

(1/2) × (22/7) × 12.25 = (1/2) × 22 × 1.75 = 11 × 1.75 = 19.25 cm²

Interactive Calculator

Length of Minute Hand (cm):

Time (minutes):

∴ The area swept by the minute hand in 30 minutes is 19.25 cm².

The length of the tangent to a circle from a point 17 cm from its Centre is 18 cm. Find the radius of the circle. (M'18)

Step 1: Apply Pythagorean Theorem

In right triangle OPT:
(OP)² = (OT)² + (PT)²
(17)² = (r)² + (18)²
289 = r² + 324

Step 2: Solve for radius

r² = 324 - 289 = 35
r = √35 ≈ 5.92 cm

Interactive Calculator

Distance from Center (cm):

Tangent Length (cm):

∴ The radius of the circle is √35 cm ≈ 5.92 cm.

Find the length of the tangent to circle from a point 13 cm away from the centre of the circle of radius 5 cm. (J'18)

Step 1: Apply Pythagorean Theorem

In right triangle OPT:
(OP)² = (OT)² + (PT)²
(13)² = (5)² + (PT)²
169 = 25 + (PT)²

Step 2: Solve for tangent length

(PT)² = 169 - 25 = 144
PT = √144 = 12 cm

Interactive Calculator

Distance from Center (cm):

Radius (cm):

∴ The length of the tangent is 12 cm.

A point P is 25 cm from the centre O of the circle. The length of the tangent drawn from P to the circle is 24 cm. Find the radius of the circle. (M'19)

Step 1: Apply Pythagorean Theorem

In right triangle OPT:
(OP)² = (OT)² + (PT)²
(25)² = (r)² + (24)²
625 = r² + 576

Step 2: Solve for radius

r² = 625 - 576 = 49
r = √49 = 7 cm

Interactive Calculator

Distance from Center (cm):

Tangent Length (cm):

∴ The radius of the circle is 7 cm.

Tangents and Secants to Circles - 4 Mark Solutions