Similar Triangles - 1 Mark Problems
Important Concepts:
• Basic Proportionality Theorem (Thales Theorem):
If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
• Midpoint Theorem:
The line segment joining midpoints of two sides is parallel to the third side and half its length.
• Similarity Criteria:
1. AAA (Angle-Angle-Angle)
2. SSS (Side-Side-Side)
3. SAS (Side-Angle-Side)
• Area Ratio:
Ratio of areas of similar triangles = (Ratio of corresponding sides)²
• Pythagorean Theorem:
In a right triangle: (Hypotenuse)² = (Base)² + (Height)²
If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
• Midpoint Theorem:
The line segment joining midpoints of two sides is parallel to the third side and half its length.
• Similarity Criteria:
1. AAA (Angle-Angle-Angle)
2. SSS (Side-Side-Side)
3. SAS (Side-Angle-Side)
• Area Ratio:
Ratio of areas of similar triangles = (Ratio of corresponding sides)²
• Pythagorean Theorem:
In a right triangle: (Hypotenuse)² = (Base)² + (Height)²
1
In a ∆DEF; A, B and C are the mid-points of EF, FD and DE respectively. If the area of ∆DEF is 14.4 cm², then find the area of ∆ABC.
(M'15)
Step 1: Apply Midpoint Theorem
When we join midpoints of sides of a triangle, the smaller triangle formed is similar to the original triangle.
Each side of ∆ABC is half the corresponding side of ∆DEF.
Step 2: Use Area Ratio Property
For similar triangles, ratio of areas = (ratio of corresponding sides)²
Ratio of sides = 1:2
Ratio of areas = (1/2)² = 1/4
Step 3: Calculate Area
Area of ∆ABC = (1/4) × Area of ∆DEF
Area of ∆ABC = (1/4) × 14.4 = 3.6 cm²
∴ The area of ∆ABC is 3.6 cm².
2
In a ∆PQR and ∆XYZ, it is given that ∆PQR ∼ ∆XYZ, ∠Y + ∠Z = 90° and XY : XZ = 3 : 4. Then find the ratio of sides in ∆PQR.
(J'15)
Step 1: Analyze given information
In ∆XYZ: ∠Y + ∠Z = 90°
Since sum of angles in a triangle = 180°
∠X = 180° - (∠Y + ∠Z) = 180° - 90° = 90°
So ∆XYZ is right-angled at X
Step 2: Apply Pythagorean Theorem
In right triangle XYZ (right-angled at X):
XY : XZ = 3 : 4
Let XY = 3k, XZ = 4k
Then YZ = √[(3k)² + (4k)²] = √[9k² + 16k²] = √25k² = 5k
Step 3: Use Similarity
Since ∆PQR ∼ ∆XYZ, corresponding sides are proportional
Ratio of sides in ∆PQR = Ratio of sides in ∆XYZ
= XY : XZ : YZ = 3k : 4k : 5k = 3 : 4 : 5
∴ The ratio of sides in ∆PQR is 3 : 4 : 5.
3
It is given that ∆ABC ∼ ∆DEF. Is it true to say that BC/CD = AB/EF? Justify your answer.
(J'16)
Step 1: Understand Similarity Property
When two triangles are similar, corresponding sides are proportional.
If ∆ABC ∼ ∆DEF, then:
AB/DE = BC/EF = AC/DF
Step 2: Analyze the given statement
The statement says: BC/CD = AB/EF
But 'D' is a vertex of ∆DEF, not a side
CD is not a side of either triangle
Step 3: Correct Proportionality
Correct proportion would be:
BC/EF = AB/DE (since BC corresponds to EF and AB corresponds to DE)
∴ No, the statement is not true. The correct proportion is BC/EF = AB/DE.
4
Draw the diagram corresponding to basic proportionality theorem.
(J'17)
Step 1: State Basic Proportionality Theorem
If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
Diagram for Basic Proportionality Theorem:
Step 2: Theorem Statement
In ∆ABC, if DE ∥ AB, then:
CD/DA = CE/EB
∴ The diagram shows triangle ABC with line DE parallel to AB, intersecting AC at D and BC at E.
5
Srivani walks 12 m due to East and turns left and walks another 5 m, how far is she from the place she started?
(M'18)
Step 1: Visualize the path
Srivani walks:
- 12 m East
- Then 5 m North (turning left from East)
Step 2: Apply Pythagorean Theorem
The path forms a right triangle with:
Base = 12 m (East)
Height = 5 m (North)
Distance from start = √(12² + 5²)
Step 3: Calculate distance
Distance = √(144 + 25) = √169 = 13 m
∴ Srivani is 13 m away from her starting point.
6
Write the similarity criterion by which pair of triangles are similar.
(M'19)
Step 1: List Similarity Criteria
There are three main criteria for triangle similarity:
1. AAA (Angle-Angle-Angle) Criterion:
If corresponding angles of two triangles are equal, then the triangles are similar.
2. SSS (Side-Side-Side) Criterion:
If corresponding sides of two triangles are proportional, then the triangles are similar.
3. SAS (Side-Angle-Side) Criterion:
If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the triangles are similar.
∴ The three similarity criteria are: AAA, SSS, and SAS.
7
Madhavi said "All squares are similar". Do you agree with her statement? Justify your answer.
(J'19)
Step 1: Definition of Similar Figures
Two figures are similar if:
1. Their corresponding angles are equal
2. Their corresponding sides are proportional
Step 2: Analyze Squares
All squares have:
- All angles equal to 90°
- All sides equal
Step 3: Check Similarity Conditions
For any two squares:
- Corresponding angles are equal (all 90°)
- Corresponding sides are proportional (ratio = side₁/side₂)
∴ Yes, I agree with Madhavi. All squares are similar because they have equal angles and proportional sides.
8
Draw a line segment of length 7.3 cm and divide it in the ratio 3 : 4.
(J'19)
Step 1: Construction Steps
To divide a line segment in ratio m:n:
Construction Method:
1. Draw line segment AB = 7.3 cm
2. Draw ray AX making an acute angle with AB
3. Mark 7 points (3+4) on AX at equal distances: A₁, A₂, A₃, A₄, A₅, A₆, A₇
4. Join A₇ to B
5. Draw A₃C ∥ A₇B, intersecting AB at C
6. Then AC:CB = 3:4
Step 2: Calculate lengths
Total parts = 3 + 4 = 7
Length of AC = (3/7) × 7.3 = 3.128... ≈ 3.13 cm
Length of CB = (4/7) × 7.3 = 4.171... ≈ 4.17 cm
∴ The line segment is divided into parts of approximately 3.13 cm and 4.17 cm in ratio 3:4.
9
The sides of a triangle measure 2√2, 4 and 2√6 units. Is it a right-angled triangle? Justify.
(Apr'23)
Step 1: Identify the longest side
Sides: 2√2 ≈ 2.828, 4, 2√6 ≈ 4.899
Longest side = 2√6 (hypotenuse if right triangle)
Step 2: Apply Pythagorean Theorem
Check if: (Longest side)² = (Side₁)² + (Side₂)²
(2√6)² = 4 × 6 = 24
(2√2)² + (4)² = (4 × 2) + 16 = 8 + 16 = 24
Step 3: Compare results
(2√6)² = 24 and (2√2)² + (4)² = 24
Since they are equal, Pythagorean theorem is satisfied
∴ Yes, it is a right-angled triangle as it satisfies the Pythagorean theorem.
10
In ∆ABC, DE is a line such that AD = 3 cm, AB = 5 cm, AE = 6 cm and AC = 10 cm. Is DE ∥ BC? Justify.
(Jun'23)
Step 1: Apply Basic Proportionality Theorem Converse
If a line divides two sides of a triangle proportionally, then it is parallel to the third side.
Step 2: Check proportionality
AD/DB = AD/(AB - AD) = 3/(5 - 3) = 3/2 = 1.5
AE/EC = AE/(AC - AE) = 6/(10 - 6) = 6/4 = 1.5
Step 3: Compare ratios
AD/DB = 1.5 and AE/EC = 1.5
Since AD/DB = AE/EC, the line DE divides sides proportionally
∴ Yes, DE ∥ BC as it divides sides AB and AC proportionally (Converse of Basic Proportionality Theorem).
Similar Triangles - 2 Marks Solutions
Important Geometric Principles:
• Pythagorean Theorem: a² + b² = c²
• Alternate Segment Theorem: Angle between tangent and chord = angle in alternate segment
• Basic Proportionality Theorem (Thales): If a line is parallel to one side, it divides other sides proportionally
• Area of Equilateral Triangle: (√3/4) × side²
• Height of Equilateral Triangle: (√3/2) × side
• Alternate Segment Theorem: Angle between tangent and chord = angle in alternate segment
• Basic Proportionality Theorem (Thales): If a line is parallel to one side, it divides other sides proportionally
• Area of Equilateral Triangle: (√3/4) × side²
• Height of Equilateral Triangle: (√3/2) × side
1
A ladder of 3.9 m length is laid against a wall. The distance between the foot of the wall and the ladder is 1.5 m. Find the height at which ladder touches the wall.
(M'15)
Step 1: Visualize the right triangle
The ladder, wall, and ground form a right triangle with:
- Hypotenuse (ladder) = 3.9 m
- Base (distance from wall) = 1.5 m
- Height (wall) = ?
- Hypotenuse (ladder) = 3.9 m
- Base (distance from wall) = 1.5 m
- Height (wall) = ?
Step 2: Apply Pythagorean Theorem
Height² + Base² = Hypotenuse²
Height² + (1.5)² = (3.9)²
Height² + 2.25 = 15.21
Height² + (1.5)² = (3.9)²
Height² + 2.25 = 15.21
Step 3: Solve for Height
Height² = 15.21 - 2.25 = 12.96
Height = √12.96 = 3.6 m
Height = √12.96 = 3.6 m
Interactive Calculator
∴ The ladder touches the wall at a height of 3.6 m.
2
AB is a chord of the circle and AOC is its diameter, such that ∠ACB = 60°. If AT is the tangent to the circle at the point A, then find the measure of ∠BAT.
(J'16)
Diagram:
Step 1: Apply Alternate Segment Theorem
The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
Step 2: Identify the relevant angles
Given: ∠ACB = 60°
By Alternate Segment Theorem:
∠BAT = ∠ACB = 60°
By Alternate Segment Theorem:
∠BAT = ∠ACB = 60°
∴ ∠BAT = 60°
3
ABC is an isosceles triangle and ∠B = 90°, then show that AC² = 2AB².
(M'16)
Step 1: Understand the triangle properties
Given: ∆ABC is isosceles with ∠B = 90°
In an isosceles triangle, two sides are equal.
Since ∠B = 90°, the equal sides must be AB and BC.
In an isosceles triangle, two sides are equal.
Since ∠B = 90°, the equal sides must be AB and BC.
Step 2: Apply Pythagorean Theorem
In right triangle ABC (right-angled at B):
AC² = AB² + BC²
Since AB = BC (isosceles triangle):
AC² = AB² + AB² = 2AB²
AC² = AB² + BC²
Since AB = BC (isosceles triangle):
AC² = AB² + AB² = 2AB²
∴ AC² = 2AB², as required.
4
In ∆ABC, PQ ‖ BC and AP = 3x - 19, PB = x - 5, AQ = x - 3, QC = 3 cm. Find x.
(M'18)
Step 1: Apply Basic Proportionality Theorem
If a line is parallel to one side of a triangle, it divides the other two sides proportionally.
So, AP/PB = AQ/QC
So, AP/PB = AQ/QC
Step 2: Substitute given values
(3x - 19)/(x - 5) = (x - 3)/3
Step 3: Cross multiply and solve
3(3x - 19) = (x - 5)(x - 3)
9x - 57 = x² - 8x + 15
0 = x² - 17x + 72
9x - 57 = x² - 8x + 15
0 = x² - 17x + 72
Step 4: Solve the quadratic equation
x² - 17x + 72 = 0
(x - 8)(x - 9) = 0
x = 8 or x = 9
(x - 8)(x - 9) = 0
x = 8 or x = 9
Solve for x
∴ The possible values of x are 8 and 9.
5
In ∆ABC, D and E are points on AB and AC respectively. If AB = 14 cm; AD = 3.5 cm, AE = 2.5 cm and AC = 10 cm, show that DE ‖ BC.
(J'18)
Step 1: Apply Converse of Basic Proportionality Theorem
If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.
Step 2: Calculate the ratios
AD/AB = 3.5/14 = 1/4
AE/AC = 2.5/10 = 1/4
AE/AC = 2.5/10 = 1/4
Step 3: Compare the ratios
Since AD/AB = AE/AC = 1/4
By the converse of Basic Proportionality Theorem, DE ‖ BC
By the converse of Basic Proportionality Theorem, DE ‖ BC
∴ DE is parallel to BC.
6
If the ratio of areas of two equilateral triangles is 25 : 36, then find the ratio of heights of the triangles.
(J'19)
Step 1: Relationship between area and side
For equilateral triangles, area = (√3/4) × side²
So, ratio of areas = (ratio of sides)²
So, ratio of areas = (ratio of sides)²
Step 2: Find ratio of sides
Given: Area₁/Area₂ = 25/36
(side₁/side₂)² = 25/36
side₁/side₂ = 5/6
(side₁/side₂)² = 25/36
side₁/side₂ = 5/6
Step 3: Find ratio of heights
Height of equilateral triangle = (√3/2) × side
So, ratio of heights = ratio of sides = 5/6
So, ratio of heights = ratio of sides = 5/6
Calculate Height Ratio
∴ The ratio of heights is 5 : 6.
Similar Triangles Construction - 4 Marks Solutions
Important Geometric Construction Principles:
• To construct a triangle similar to a given triangle with a given scale factor
• Steps: 1) Draw the original triangle, 2) Draw a ray making an acute angle with one side
• Mark points on the ray according to the scale factor, 3) Join the last point to the vertex
• Draw lines parallel to this line through the other points
• Steps: 1) Draw the original triangle, 2) Draw a ray making an acute angle with one side
• Mark points on the ray according to the scale factor, 3) Join the last point to the vertex
• Draw lines parallel to this line through the other points
1
Construct a triangle of sides 5cm, 6cm and 7cm then construct a triangle similar to it, whose sides are 2/3 of the corresponding sides of the first triangle.
(M'16)
Step 1: Construct the original triangle
Draw BC = 7 cm
With B as center and radius 6 cm, draw an arc
With C as center and radius 5 cm, draw another arc intersecting the first arc at A
Join AB and AC to get ΔABC
With B as center and radius 6 cm, draw an arc
With C as center and radius 5 cm, draw another arc intersecting the first arc at A
Join AB and AC to get ΔABC
Step 2: Draw a ray BX making an acute angle with BC
Mark 3 equal points B₁, B₂, B₃ on BX such that BB₁ = B₁B₂ = B₂B₃
Step 3: Join B₃ to C
Draw a line through B₂ parallel to B₃C, intersecting BC at C'
Step 4: Draw line through C' parallel to AC
This line intersects AB at A'
ΔA'BC' is the required triangle with sides 2/3 of ΔABC
ΔA'BC' is the required triangle with sides 2/3 of ΔABC
Diagram:
∴ ΔA'BC' is the required triangle with sides 2/3 of the original triangle.
2
Construct a triangle of sides 5cm, 6cm and 7cm then construct a triangle similar to it, whose sides are 1½ times the corresponding sides of the first triangle.
(J'16)
Step 1: Construct the original triangle
Draw BC = 7 cm
With B as center and radius 6 cm, draw an arc
With C as center and radius 5 cm, draw another arc intersecting the first arc at A
Join AB and AC to get ΔABC
With B as center and radius 6 cm, draw an arc
With C as center and radius 5 cm, draw another arc intersecting the first arc at A
Join AB and AC to get ΔABC
Step 2: Draw a ray BX making an acute angle with BC
Mark 2 equal points B₁, B₂ on BX such that BB₁ = B₁B₂
Since scale factor is 3/2, we need to mark 2 points
Since scale factor is 3/2, we need to mark 2 points
Step 3: Join B₂ to C
Draw a line through B₁ parallel to B₂C, intersecting BC extended at C'
Step 4: Draw line through C' parallel to AC
This line intersects AB extended at A'
ΔA'BC' is the required triangle with sides 3/2 of ΔABC
ΔA'BC' is the required triangle with sides 3/2 of ΔABC
∴ ΔA'BC' is the required triangle with sides 1½ times the original triangle.
3
Construct an equilateral triangle XYZ of side 5 cm and construct another triangle similar to triangle XYZ, each of its sides is 4/5 of the sides of ΔXYZ.
(M'18)
Step 1: Construct the equilateral triangle
Draw XY = 5 cm
With X as center and radius 5 cm, draw an arc
With Y as center and radius 5 cm, draw another arc intersecting the first arc at Z
Join XZ and YZ to get equilateral ΔXYZ
With X as center and radius 5 cm, draw an arc
With Y as center and radius 5 cm, draw another arc intersecting the first arc at Z
Join XZ and YZ to get equilateral ΔXYZ
Step 2: Draw a ray XW making an acute angle with XY
Mark 5 equal points X₁, X₂, X₃, X₄, X₅ on XW such that XX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅
Step 3: Join X₅ to Y
Draw a line through X₄ parallel to X₅Y, intersecting XY at Y'
Step 4: Draw line through Y' parallel to YZ
This line intersects XZ at Z'
ΔXY'Z' is the required triangle with sides 4/5 of ΔXYZ
ΔXY'Z' is the required triangle with sides 4/5 of ΔXYZ
∴ ΔXY'Z' is the required triangle with sides 4/5 of the original equilateral triangle.
4
Construct a triangle ABC in which AB = 5 cm, BC = 7 cm and angle B = 50°, then construct a triangle similar to it, whose sides are 4/5 of the corresponding sides of first triangle.
(J'18)
Step 1: Construct the original triangle
Draw BC = 7 cm
At B, construct ∠CBX = 50°
With B as center and radius 5 cm, cut BX at A
Join AC to get ΔABC
At B, construct ∠CBX = 50°
With B as center and radius 5 cm, cut BX at A
Join AC to get ΔABC
Step 2: Draw a ray BY making an acute angle with BC
Mark 5 equal points B₁, B₂, B₃, B₄, B₅ on BY such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
Step 3: Join B₅ to C
Draw a line through B₄ parallel to B₅C, intersecting BC at C'
Step 4: Draw line through C' parallel to AC
This line intersects AB at A'
ΔA'BC' is the required triangle with sides 4/5 of ΔABC
ΔA'BC' is the required triangle with sides 4/5 of ΔABC
∴ ΔA'BC' is the required triangle with sides 4/5 of the original triangle.
5
Construct a triangle PQR, in which PQ = 4 cm, QR = 6 cm and ∠PQR = 70°. Construct triangle such that each side of the new triangle is 3/4 of the triangle PQR.
(M'19)
Step 1: Construct the original triangle
Draw PQ = 4 cm
At Q, construct ∠PQR = 70°
With Q as center and radius 6 cm, cut QR at R
Join PR to get ΔPQR
At Q, construct ∠PQR = 70°
With Q as center and radius 6 cm, cut QR at R
Join PR to get ΔPQR
Step 2: Draw a ray QX making an acute angle with PQ
Mark 4 equal points Q₁, Q₂, Q₃, Q₄ on QX such that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄
Step 3: Join Q₄ to P
Draw a line through Q₃ parallel to Q₄P, intersecting QP at P'
Step 4: Draw line through P' parallel to PR
This line intersects QR at R'
ΔP'QR' is the required triangle with sides 3/4 of ΔPQR
ΔP'QR' is the required triangle with sides 3/4 of ΔPQR
∴ ΔP'QR' is the required triangle with sides 3/4 of the original triangle.
6
In a right-angle triangle, the hypotenuse is 10 cm more than the shortest side. If third side is 6 cm less than the hypotenuse, find the sides of the right-angle triangle.
(M'19)
Step 1: Define variables
Let shortest side = x cm
Then hypotenuse = (x + 10) cm
Third side = (x + 10 - 6) = (x + 4) cm
Then hypotenuse = (x + 10) cm
Third side = (x + 10 - 6) = (x + 4) cm
Step 2: Apply Pythagorean Theorem
x² + (x + 4)² = (x + 10)²
x² + x² + 8x + 16 = x² + 20x + 100
2x² + 8x + 16 = x² + 20x + 100
x² + x² + 8x + 16 = x² + 20x + 100
2x² + 8x + 16 = x² + 20x + 100
Step 3: Simplify the equation
x² - 12x - 84 = 0
Step 4: Solve the quadratic equation
Using quadratic formula:
x = [12 ± √(144 + 336)] / 2
x = [12 ± √480] / 2
x = [12 ± 4√30] / 2
x = 6 ± 2√30
Taking positive value: x = 6 + 2√30 ≈ 6 + 10.95 = 16.95 cm
x = [12 ± √(144 + 336)] / 2
x = [12 ± √480] / 2
x = [12 ± 4√30] / 2
x = 6 ± 2√30
Taking positive value: x = 6 + 2√30 ≈ 6 + 10.95 = 16.95 cm
Calculate Triangle Sides
∴ The sides are: Shortest side ≈ 16.95 cm, Third side ≈ 20.95 cm, Hypotenuse ≈ 26.95 cm.
7
In a right-angle triangle, length of the hypotenuse is 6 cm more than its shortest side. The length of the other side is 3 cm less than the hypotenuse, then find the sides of right-angle triangle.
(J'19)
Step 1: Define variables
Let shortest side = x cm
Then hypotenuse = (x + 6) cm
Third side = (x + 6 - 3) = (x + 3) cm
Then hypotenuse = (x + 6) cm
Third side = (x + 6 - 3) = (x + 3) cm
Step 2: Apply Pythagorean Theorem
x² + (x + 3)² = (x + 6)²
x² + x² + 6x + 9 = x² + 12x + 36
2x² + 6x + 9 = x² + 12x + 36
x² + x² + 6x + 9 = x² + 12x + 36
2x² + 6x + 9 = x² + 12x + 36
Step 3: Simplify the equation
x² - 6x - 27 = 0
Step 4: Solve the quadratic equation
x² - 6x - 27 = 0
(x - 9)(x + 3) = 0
x = 9 or x = -3 (discard negative value)
So x = 9 cm
(x - 9)(x + 3) = 0
x = 9 or x = -3 (discard negative value)
So x = 9 cm
Calculate Triangle Sides
∴ The sides are: Shortest side = 9 cm, Third side = 12 cm, Hypotenuse = 15 cm.
8
Construct triangle ABC with BC = 7cm, angle B = 45° and angle C = 60°. Then construct another triangle similar to ΔABC, whose sides are 3/5 times of the corresponding sides of ΔABC.
(Apr'23)
Step 1: Construct the original triangle
Draw BC = 7 cm
At B, construct ∠CBX = 45°
At C, construct ∠BCY = 60° such that it intersects BX at A
Join AB and AC to get ΔABC
At B, construct ∠CBX = 45°
At C, construct ∠BCY = 60° such that it intersects BX at A
Join AB and AC to get ΔABC
Step 2: Draw a ray BZ making an acute angle with BC
Mark 5 equal points B₁, B₂, B₃, B₄, B₅ on BZ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
Step 3: Join B₅ to C
Draw a line through B₃ parallel to B₅C, intersecting BC at C'
Step 4: Draw line through C' parallel to AC
This line intersects AB at A'
ΔA'BC' is the required triangle with sides 3/5 of ΔABC
ΔA'BC' is the required triangle with sides 3/5 of ΔABC
∴ ΔA'BC' is the required triangle with sides 3/5 of the original triangle.
