In the following figure- - ABC-90-AB - x-8 cm- BC- x-1 cm and AC-x - 15 cm-Find the value of x-

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Pythagoras Theorem

In the follow...

Question

In the following figure, $∠ A B C = 90^{\circ}, A B = (x + 8) c m, B C = (x + 1) c m$ and $A C = (x + 15) c m$ .
Find the value of $x$ .

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Solution

In a right angled triangle, ${Hypotenuse}^{2} = {base}^{2} + {height}^{2}$
So, ${A C}^{2} = {A B}^{2} + {B C}^{2}$
$\Rightarrow {(x + 15)}^{2} = {(x + 8)}^{2} + {(x + 1)}^{2}$
$\Rightarrow x^{2} + 225 + 30 x = x^{2} + 64 + 16 x + x^{2} + 1 + 2 x$
$\Rightarrow 225 + 30 x = x^{2} + 65 + 18 x$
$\Rightarrow x^{2} - 160 - 12 x = 0$
$\Rightarrow x^{2} - 20 x + 8 x - (20 \times 8) = 0$
$\Rightarrow x (x - 20) + 8 (x - 20) = 0$
$\Rightarrow (x - 20) (x + 8) = 0$
$\Rightarrow x = 20$ or $x = - 8$
Since, length cannot be negative,
$\Rightarrow x = 20$

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In the following figure, ∠ABC=90∘,AB=(x+8) cm,BC=(x+1) cm and AC=(x+15) cm.Find the value of x.

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In the following figure, $∠ A B C = 90^{\circ}, A B = (x + 8) c m, B C = (x + 1) c m$ and $A C = (x + 15) c m$ .
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