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Practice Set 2.2 | Q 4 | Page 43
In ∆ABC, point M is the midpoint of side BC.
If, AB2 + AC2 = 290 cm2, AM = 8 cm, find BC.

In D ABC, point M is the midpoint of side BC.
Solution
\[BM = MC = \frac{1}{2}BC\]

\[{AB}^2 + {AC}^2 = 2 {AM}^2 + 2 {BM}^2 \left( \text{by Apollonius theorem} \right)\]
\[ \Rightarrow 290 = 2 \left( 8 \right)^2 + 2 {BM}^2 \]
\[ \Rightarrow 290 = 2\left( 64 \right) + 2 {BM}^2 \]
\[ \Rightarrow 290 = 128 + 2 {BM}^2 \]
\[ \Rightarrow 2 {BM}^2 = 290 - 128\]
\[ \Rightarrow 2 {BM}^2 = 162\]
\[ \Rightarrow {BM}^2 = 81\]
\[ \Rightarrow BM = 9\]
\[ \therefore BC = 2 \times BM\]
\[ = 2 \times 9\]
\[ = 18 cm\]

Hence, BC = 18 cm.

Explanation:- 

Given: In ∆ABC, BM=MC=1/2BC and AB²+AC²=2AM²+2BM²

To Find: BC

Proof:

Using BM=MC=1/2BC, we can say that BM=MC=BC/2.

Let’s apply Apollonius theorem for side AB,

AB²+AC²=2AM²+2BM²

Given AB=8, AC=√290 and AM=4 (as M is the midpoint of BC)

Substituting the given values in the above formula, we get:

8²+ (√290)²=2(4)²+2BM²

Simplifying the above equation, we get:

BM²=81

Taking square root of both sides, we get:

BM=9

As we know, BM=MC=BC/2. Substituting BM=MC=9, we get:

9+9=BC/2

18=BC/2

Multiplying both sides by 2, we get:

BC=18

Therefore, the length of BC is 18 cm.

Chapter 2 – Pythagoras Theorem- Text Book Solution

Practice Set 2.2 | Q  4 | Page 43