The difference between the sides at right angle in a right-angled… - Vidyadip

Question

The difference between the sides at right angle in a right-angled triangle is 7cm. The area of the triangle is $60cm^2$. Find its perimeter.

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Answer

Given,
Area of the triangle = $60cm^2$
Let the side of the triangle be a, b and c, where a is the height, b is the base and c is hypotense of the triangle.
a - b = 7cm
a = 7 + b ...(1)
Area of triangle $=\frac{1}{2}\times\text{b}\times\text{h}$
$\Rightarrow6=\frac12\times\text{b}\times(7+\text{b})$
$\Rightarrow120=7\text{b}+\text{b}^2$
$\Rightarrow\text{b}^2+7\text{b}-12=0$
$\Rightarrow(\text{b}+15)(\text{b}-8)=0$
$\Rightarrow\text{b}=-15$ or 8
Side of a triangle cannot be negative.
Threfore, b = 8cm.
Substituting the value of b = 8cm, in equatin (1):
a = 7 + 8 = 15cm
Now, a = 15cm, b = 8cm
Now, in the given right triangle, we have to find third side.
$(Hyp)^2 = (First side)^2 + (Second side)^2$
$\Rightarrow Hyp^2 = 8^2 +15^2$
$\Rightarrow Hyp^2 = 64 + 225$
$\Rightarrow Hyp^2 = 289$
$\Rightarrow Hyp = 17cm$
So, the third side is 17cm.
Perimeter of a triangle = a + b + c.
$\therefore$ required perimeter of the triangle = 15 + 8 + 17 = 40cm.

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