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 triangle $= 24cm^2$
Let the sides be a and b, where a is the height and b is the base of triangle.
a - b = 2cm a - b = 2cm
a = 2 + b ...(1)
Area of triangle $=\frac{1}{2}\times\text{b}\times\text{h}$
$\Rightarrow24=\frac12\times\text{b}\times(2+\text{b})$
$\Rightarrow48=\text{b}+\frac12\text{b}^2$
$\Rightarrow48=2\text{b}+\text{b}^2$
$\Rightarrow\text{b}^2+2\text{b}-48=0$
$\Rightarrow(\text{b}+8)(\text{b}-6)=0$
$\Rightarrow\text{b}=-8$ or 6
Side of a triangle cannot be negative.
Threfore, b = 6cm.
Substituting the value of b = 6cm, in equatin (1) we get:
a = 2 + 6 = 8cm
Now, a = 8cm, b = 6cm
Now, in the given right triangle, we have to find third side.
Using the relation
$(Hyp)^2 = (Oneside)^2 + (Otherside)^2$
$\Rightarrow Hyp^2 = 8^2 + 6^2$
$\Rightarrow Hyp^2 = 64 + 36$
$\Rightarrow Hyp^2 = 100$
$\Rightarrow Hyp = 10cm$
So, the third side is 10cm.
So, Perimeter of a triangle = a + b + c.
$\therefore$ required perimeter of the triangle = 8 + 6 + 10 = 24cm.

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