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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.

\[AD = BD = \frac{1}{2}AB = 5\]

\[{CA}^2 + {CB}^2 = 2 {DC}^2 + 2 {AD}^2 \left( \text{by Apollonius theorem} \right)\]

\[ \Rightarrow 7^2 + 9^2 = 2 {DC}^2 + 2\left( 5^2 \right)\]

\[ \Rightarrow 49 + 81 = 2 {DC}^2 + 2\left( 25 \right)\]

\[ \Rightarrow 130 = 2 {DC}^2 + 50\]

\[ \Rightarrow 2 {DC}^2 = 130 - 50\]

\[ \Rightarrow 2 {DC}^2 = 80\]

\[ \Rightarrow {DC}^2 = 40\]

\[ \Rightarrow DC = 2\sqrt{10}\]

Hence, the length of the median drawn from point D to side AB is \[2\sqrt{10}\]

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 + 2\left( 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 = 2\sqrt{10}]

Thus, the length of the median drawn from point D to side AB is equal to the length of DC, which is [2\sqrt{10}].

Chapter 2 – Pythagoras Theorem- Text Book Solution

Practice Set 2.2 | Q  2 | Page 43