The difference between the sides at right angle in a right-angled triangle is 7 cm. The area of the triangle is 60 cm2. Find its perimeter.
Given:
Area of the triangle $=60 \mathrm{~cm}^{2}$
Let the sides of the triangle be a, b and c, where a is the height, b is the base and c is hypotenuse of the triangle.
$a-b=7 \mathrm{~cm}$
$a=7+b \ldots \ldots .(1)$
Area of triangle $=\frac{1}{2} \times b \times h$
$\Rightarrow 60=\frac{1}{2} \times b \times(7+b)$
$\Rightarrow 120=7 b+b^{2}$
$\Rightarrow b^{2}+7 b-120=0$
$\Rightarrow(b+15)(b-8)=0$
$\Rightarrow b=-15$ or 8
Side of a triangle cannot be negative.
Therefore, b = 8 cm.
Substituting the value of b = 8 cm, in equation (1):
a = 7+8 = 15 cm
Now, a = 15 cm, b = 8 cm
Now, in the given right triangle, we have to find third side.
$(\mathrm{Hyp})^{2}=(\text { First side })^{2}+(\text { Second side })^{2}$
$\Rightarrow \mathrm{Hyp}^{2}=8^{2}+15^{2}$
$\Rightarrow \mathrm{Hyp}^{2}=64+225$
$\Rightarrow \mathrm{Hyp}^{2}=289$
$\Rightarrow \mathrm{Hyp}=17 \mathrm{~cm}$
So, the third side is 17 cm
Perimeter of a triangle $=a+b+c$.
Therefore, required perimeter of the triangle $=15+8+17=40 \mathrm{~cm}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.