# ISI2015-MMA-28

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In a triangle $ABC$, $AD$ is the median. If length of $AB$ is $7$, length of $AC$ is $15$ and length of $BC$ is $10$ then length of $AD$ equals

1. $\sqrt{125}$
2. $69/5$
3. $\sqrt{112}$
4. $\sqrt{864}/5$

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@ankitgupta.1729

Any idea how this can be solved??

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Answer: $\mathbf C$ Given:

$AC = 15$
$BC = 10$

$\because$ $D$ is the median.

$\therefore BD = CD = 5$

Using $\text{Cosine Rule}$

$$\cos B = \frac{a^2+c^2-b^2}{2ac}$$

$$\therefore \cos B = \frac{ AB^2 + BC^2 -2AC^2}{2AB\times BC}$$

$$\implies \cos B = \frac{49 + 100 -15^2}{140}$$

$$\implies \cos B = \frac{-19}{35}$$

Now, we have:

$$AD^2 = AB^2 + BD^2 -2AB\times BD \cos B$$

$$\implies AD^2 = 49+25-2\times7\times5\bigg(\frac{-19}{35}\bigg)$$

$$\implies AD^2 = 112$$

$$\implies AD = \sqrt{112}$$

$\therefore\;\bf{C}$ is the correct option.

edited by

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