Practice Set 2.2 | Q 2 | Page 43
In ∆ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB
Solution
In ∆ACB, point D is the midpoint of side AB.
[AD = BD = frac{1}{2}AB = 5]
[{CA}^2 + {CB}^2 = 2 {DC}^2 + 2 {AD}^2 left( text{by Apollonius theorem} right)]
[ Rightarrow 7^2 + 9^2 = 2 {DC}^2 + 2left( 5^2 right)]
[ Rightarrow 49 + 81 = 2 {DC}^2 + 2left( 25 right)]
[ Rightarrow 130 = 2 {DC}^2 + 50]
[ Rightarrow 2 {DC}^2 = 130 – 50]
[ Rightarrow 2 {DC}^2 = 80]
[ Rightarrow {DC}^2 = 40]
[ Rightarrow DC = 2sqrt{10}]
Hence, the length of the median drawn from point D to side AB is [2sqrt{10}]
Explanation:-
Given a triangle ∆ACB where D is the midpoint of AB, it can be seen that AD = BD = AB/2 = 5.
Using the Apollonius theorem, we know that:
[{CA}^2 + {CB}^2 = 2 {DC}^2 + 2 {AD}^2 ]
Substituting the given values, we have:
[ 7^2 + 9^2 = 2 {DC}^2 + 2left( 5^2 right)]
Simplifying, we get:
[ 49 + 81 = 2 {DC}^2 + 2(25)] [130 = 2 {DC}^2 + 50] [2 {DC}^2 = 80] [{DC}^2 = 40] [DC = 2sqrt{10}]
Thus, the length of the median drawn from point D to side AB is equal to the length of DC, which is [2sqrt{10}].
Chapter 2 – Pythagoras Theorem- Text Book Solution
Practice Set 2.2 | Q 2 | Page 43