In $Δ A B C$ , $∠ B = 90^{\circ}$ , Find the sides of the triangle, if $A B = x c m, B C = (4 x + 4) c m$ and $A C = (4 x + 5) c m$

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Solution

In a right angled triangle with right angle at $B$ ,
$A B$ and $B C$ for the sides and $A C$ is the hypotenuse.

So, as per the pythagoras theorem, ${A C}^{2} = {A B}^{2} + {B C}^{2}$
$\Rightarrow {(4 x + 5)}^{2} = {(x)}^{2} + {(4 x + 4)}^{2}$
$\Rightarrow 16 x^{2} + 40 x + 25 = x^{2} + 16 x^{2} + 32 x + 16$
$\Rightarrow x^{2} - 8 x - 9 = 0$
$\Rightarrow x^{2} - 9 x + x - 9 = 0$
$\Rightarrow x (x - 9) + 1 (x - 11) = 0$
$\Rightarrow x = - 1, x = 9$

When $x = - 1,$ side $A B = - 1$ which is not possible.

Hence, $x = 9$ and sides are $A C = 4 x + 5 = 41; A B = x = 9; B C = 4 x + 4 = 40$

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