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Find x in the given Right angled triangle
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Given: Right-angled triangle

To find: $x$

Solution:

According to the Pythagoras Theorem formula :


Hypotenuse2 = Perpendicular2 + Base2 

(x+2)2 = 52 + (x-3)2 

$(x+2)^{2}=5^{2}+(x-3)^{2}$

$5^{2}=25$

$(x+2)^{2}=25+(x-3)^{2}$

Expand $(x+2)^{2}: x^{2}+4 x+4$

Expand $25+(x-3)^{2}: \quad x^{2}-6 x+34$

$x^{2}+4 x+4=x^{2}-6 x+34$

Subtract 4 from both sides

$x^{2}+4 x+4-4=x^{2}-6 x+34-4$

Simplify
$x^{2}+4 x=x^{2}-6 x+30$

Subtract $x^{2}-6 x$ from both sides
$x^{2}+4 x-\left(x^{2}-6 x\right)=x^{2}-6 x+30-\left(x^{2}-6 x\right)$

Simplify
$10 x=30$

Divide both sides by 10
$\frac{10 x}{10}=\frac{30}{10}$

Simplify
$x=3$

Updated on: 10-Oct-2022

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