Question

Find the radius of the incircle of a triangle whose sides are 15 , 20, 25

Answer 1

Solution:

In-circle of any triangle  is the point where internal bisectors of angles of Triangle Meet.

To draw in-circle of triangle whose sides are 15 units, 20 units and 25 units we need to first construct he triangle.

Draw a line segment AB of length of length 15 cm.

From B mark an arc of length 20 units from A either above line AB or below line AB.

From A mark an arc cutting the arc marked from Point B at a distance of 25 units from A.You will get point C.

This is Required Triangle ABC.

To draw the in-circle of triangle ABC , draw internal bisector of Angle A as well as angle B or angle C. The point where these internal bisectors of angles Intersects is in-center of triangle ABC. Locate this as Point O. From O join O M , ON .Taking Any one of them as radius Draw in-circle of triangle ABC.

Area of Triangle ABC = $\sqrt{S(S-a)(S-b)(S-c)}, where S= \frac{a+b+c}{2}\\\\\ S= \frac{25 +15+20}{2}\\\\ S= 30 {\text{units}}\\Area of Triangle ABC = \sqrt{(30-25)(30-20)(30-15)}\\\\ \sqrt(150)^2=150 {\text{square units}}$-----(1)

Let OP= OQ=OR= r

Area (Δ ABC)= $\frac{1}{2} \times r \times 25 + \frac{1}{2} \times r \times 20+\frac{1}{2} \times r \times 15=\frac{1}{2} \times r \times 60= 30 r$-----(2)

Equating (1) and (2)

30 r = 150

Dividing both sides by 30, we get

r= 5 units