# Please solve q 95 ?

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May 12, 2018

The length of the longest side is $21$.

#### Explanation:

In a $\Delta A B C$,

$\rightarrow \cos A = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}$

$\rightarrow A r e a = \left(\frac{1}{2}\right) a \cdot b \sin C$

Now, $A r e a$ of $\Delta A B D = \left(\frac{1}{2}\right) \cdot 9 \cdot 8 \cdot \sin x = 36 \sin x$

$A r e a$ of $\Delta A D C = \left(\frac{1}{2}\right) \cdot 8 \cdot 18 \cdot \sin x = 72 \sin x$

$A r e a$ of $\Delta A B C = \left(\frac{1}{2}\right) \cdot 9 \cdot 18 \cdot \sin 2 x = 81 \sin 2 x$

$\rightarrow \Delta A B C = \Delta A B D + \Delta A D C$

$\rightarrow 81 \sin 2 x = 36 \cdot \sin x + 72 \cdot \sin x = 108 \cdot \sin x$

$\rightarrow 81 \cdot 2 \cancel{\sin x} \cdot \cos x = 108 \cdot \cancel{\sin x}$

$\rightarrow \cos x = \frac{108}{162} = \frac{2}{3}$

Applying cosine law in $\Delta A B C$, we get,

$\rightarrow \cos 2 x = \frac{{9}^{2} + {18}^{2} - {a}^{2}}{2 \cdot 9 \cdot 18}$

$\rightarrow 2 {\cos}^{2} x - 1 = \frac{405 - {a}^{2}}{324}$

$\rightarrow 2 \cdot {\left(\frac{2}{3}\right)}^{2} - 1 = \frac{405 - {a}^{2}}{324}$

$\rightarrow 2 \cdot \left(\frac{4}{9}\right) - 1 = \frac{405 - {a}^{2}}{324}$

$\rightarrow - 36 = 405 - {a}^{2}$

$\rightarrow {a}^{2} = 405 + 36 = 441$

$\rightarrow a = 21$

Also, note that

$\rightarrow \sin 2 x = 2 \sin x \cos x$

$\rightarrow \cos 2 x = 2 {\cos}^{2} x - 1$