Table of Contents
Exercise 8.3 Solutions
Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad
Problem 1
D, E, F are mid points of sides BC, CA, AB of \(\triangle ABC\). Find the ratio of areas of \(\triangle DEF\) and \(\triangle ABC\).
Solution:
Since D, E, F are midpoints:
DE = ½AB, EF = ½BC, FD = ½AC (by midpoint theorem)
Thus, \(\triangle DEF \sim \triangle ABC\) with similarity ratio 1:2
Ratio of areas = (ratio of sides)² = (½)² = ¼
Therefore, \(\frac{\text{Area of } \triangle DEF}{\text{Area of } \triangle ABC} = \frac{1}{4}\)
Problem 2
In \(\triangle ABC\), XY \(\parallel\) AC and XY divides the triangle into two parts of equal area. Find the ratio of \(\frac{AX}{XB}\).
Solution:
Since XY ∥ AC, \(\triangle BXY \sim \triangle BAC\) by AA similarity
Let area of \(\triangle ABC = 2A\), then area of \(\triangle BXY = A\)
Thus, \(\frac{\text{Area } \triangle BXY}{\text{Area } \triangle BAC} = \frac{1}{2}\)
But ratio of areas = (ratio of sides)² ⇒ \(\left(\frac{BX}{BA}\right)^2 = \frac{1}{2}\)
\(\Rightarrow \frac{BX}{BA} = \frac{1}{\sqrt{2}}\)
\(\Rightarrow \frac{AX}{XB} = \frac{BA – BX}{BX} = \frac{\sqrt{2} – 1}{1} = \sqrt{2} – 1\)
Rationalizing: \(\frac{AX}{XB} = \frac{\sqrt{2} – 1}{1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{2 – 1}{\sqrt{2} + 1} = \frac{1}{\sqrt{2} + 1}\)
Problem 3
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Let \(\triangle ABC \sim \triangle DEF\) with ratio of similarity k:1
Let AM and DN be corresponding medians
Since corresponding sides and medians are proportional in similar triangles:
\(\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = \frac{AM}{DN} = k\)
Ratio of areas = \(\left(\frac{AB}{DE}\right)^2 = k^2\)
But \(\left(\frac{AM}{DN}\right)^2 = k^2\)
Thus, \(\frac{\text{Area } \triangle ABC}{\text{Area } \triangle DEF} = \left(\frac{AM}{DN}\right)^2\)
Problem 4
\(\triangle ABC \sim \triangle DEF\). BC = 3cm, EF = 4cm and area of \(\triangle ABC = 54 \, \text{cm}^2\). Determine the area of \(\triangle DEF\).
Solution:
Ratio of corresponding sides = \(\frac{BC}{EF} = \frac{3}{4}\)
Ratio of areas = (ratio of sides)² = \(\left(\frac{3}{4}\right)^2 = \frac{9}{16}\)
Let area of \(\triangle DEF = x\)
\(\frac{54}{x} = \frac{9}{16}\)
\(\Rightarrow x = \frac{54 \times 16}{9} = 96 \, \text{cm}^2\)
Problem 5
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, BP = 3cm, AQ = 1.5 cm and CQ = 4.5 cm, prove that area of \(\triangle APQ = \frac{1}{16}\) (area of \(\triangle ABC\)).
Solution:
Given: AP = 1 cm, BP = 3 cm ⇒ AB = AP + BP = 4 cm
AQ = 1.5 cm, CQ = 4.5 cm ⇒ AC = AQ + CQ = 6 cm
In \(\triangle APQ\) and \(\triangle ABC\):
\(\angle A\) is common
\(\frac{AP}{AB} = \frac{1}{4}\), \(\frac{AQ}{AC} = \frac{1.5}{6} = \frac{1}{4}\)
Thus, \(\triangle APQ \sim \triangle ABC\) by SAS similarity with ratio 1:4
Ratio of areas = (1:4)² = 1:16
Therefore, \(\text{Area of } \triangle APQ = \frac{1}{16} \text{Area of } \triangle ABC\)
Problem 6
The areas of two similar triangles are \(81 \, \text{cm}^2\) and \(49 \, \text{cm}^2\) respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.
Solution:
Ratio of areas = \(\frac{81}{49} = \left(\frac{9}{7}\right)^2\)
Thus, ratio of corresponding altitudes = \(\frac{9}{7}\)
Let altitude of smaller triangle = h
\(\frac{4.5}{h} = \frac{9}{7}\)
\(\Rightarrow h = \frac{4.5 \times 7}{9} = 3.5 \, \text{cm}\)
