In $ \triangle \mathrm{ABC}, \mathrm{AD} $ is a median. If $ \mathrm{AB}=8, \mathrm{AC}=15 $ and $ \mathrm{AD}=8.5 $, find $ \mathrm{BC} $.

Given:

In \( \triangle \mathrm{ABC}, \mathrm{AD} \) is a median.

\( \mathrm{AB}=8, \mathrm{AC}=15 \) and \( \mathrm{AD}=8.5 \)
To do:

We have to find \( \mathrm{BC} \).

Solution:

We know that,

The sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side.

Therefore,

$AB^2 + AC^2 = 2(AD^2 + BD^2)$

$8^2 + 15^2 = 2(8.5^2 + BD^2)$

$64 + 225 = 2(72.25 + BD^2)$

$289 = 144.5 + 2BD^2$

$2BD^2 = 289 - 144.5$

$2BD^2 = 144.5$

$BD^2 = \frac{144.5}{2}=72.25$

$BD^2 = (8.5)^2$

$\Rightarrow BD = 8.5$

AD is a median, this implies,

$\frac{BC}{2} = BD = DC = 8.5$

$\Rightarrow BC = 8.5 \times 2 = 17$

Hence, $BC=17$.

Updated on: 10-Oct-2022

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