#### Question

In ∆ABC, point M is the midpoint of side BC.

If, AB^{2 }+ AC^{2 }= 290 cm^{2}, AM = 8 cm, find BC.

#### Solution

In ∆ABC, point M is the midpoint of side BC.

\[BM = MC = \frac{1}{2}BC\]

\[{AB}^2 + {AC}^2 = 2 {AM}^2 + 2 {BM}^2 \left( \text{by Apollonius theorem} \right)\]

\[ \Rightarrow 290 = 2 \left( 8 \right)^2 + 2 {BM}^2 \]

\[ \Rightarrow 290 = 2\left( 64 \right) + 2 {BM}^2 \]

\[ \Rightarrow 290 = 128 + 2 {BM}^2 \]

\[ \Rightarrow 2 {BM}^2 = 290 - 128\]

\[ \Rightarrow 2 {BM}^2 = 162\]

\[ \Rightarrow {BM}^2 = 81\]

\[ \Rightarrow BM = 9\]

\[ \therefore BC = 2 \times BM\]

\[ = 2 \times 9\]

\[ = 18 cm\]

Hence, BC = 18 cm.

Is there an error in this question or solution?

Solution In ∆Abc, Point M is the Midpoint of Side Bc. If, Ab2 + Ac2 = 290 Cm2, Am = 8 Cm, Find Bc. Concept: Apollonius Theorem.