Question
The difference between the sides at right angles in a right$-$angled triangle is $14 \ cm .$ The area of the triangle is $120 \ cm^2$. Calculate the perimeter of the triangle.
✓
Answer
Given that, the difference between the sides at right angles in a right$-$angled triangle is $14 \ cm .$
Let the sides containing the right angle be $x \ cm$ and $( x -14 ) \ cm$
Then, the area of the triangle $=\left[\frac{1}{2} \times x \times(x-14)\right] \ cm ^2$
But, area $=120 \ cm^2 ($ given $).$
$\therefore \frac{1}{2} x(x-14)=120$
$\Rightarrow x^2-14 x-240=0$
$\Rightarrow x^2-24 x+10 x-240$
$\Rightarrow x(x-24)+10(x-24)$
$\Rightarrow(x-24)(x+10)=0$
$\Rightarrow x=24 ($ neglecting $x=-10)$
$\therefore$ one side $=24 \ cm$ other side $=(24-14) \ cm=10 \ cm$
Hypotenuse $=\sqrt{(24)^2+(10)^2} \ cm$
$=\sqrt{576+100} \ cm$
$=\sqrt{676} \ cm$
$=26 \ cm$
$\therefore$ perimeter of the triangle $=(24+10+26) \ cm =60 \ cm$.
Let the sides containing the right angle be $x \ cm$ and $( x -14 ) \ cm$
Then, the area of the triangle $=\left[\frac{1}{2} \times x \times(x-14)\right] \ cm ^2$
But, area $=120 \ cm^2 ($ given $).$
$\therefore \frac{1}{2} x(x-14)=120$
$\Rightarrow x^2-14 x-240=0$
$\Rightarrow x^2-24 x+10 x-240$
$\Rightarrow x(x-24)+10(x-24)$
$\Rightarrow(x-24)(x+10)=0$
$\Rightarrow x=24 ($ neglecting $x=-10)$
$\therefore$ one side $=24 \ cm$ other side $=(24-14) \ cm=10 \ cm$
Hypotenuse $=\sqrt{(24)^2+(10)^2} \ cm$
$=\sqrt{576+100} \ cm$
$=\sqrt{676} \ cm$
$=26 \ cm$
$\therefore$ perimeter of the triangle $=(24+10+26) \ cm =60 \ cm$.
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