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Chapter 8: Q. 53 (page 543)

The lines that bisect each angle of a triangle meet in a single point O, and the perpendicular distance r from O to each side of the triangle is the same. The circle with center at O and radius r is called the inscribed circle of the triangle.

Show that the area K of triangle PQR is K=rs, where, s=12a+b+c.

Then show thatr=s-as-bs-cs.

Short Answer

Expert verified

To show that r=s-as-bs-cs, first find the area of the three triangles and add them, followed by simplification.

Step by step solution

01

Step 1. Given information.

Consider the given question,

r=s-as-bs-cs......(i)s=12a+b+c......(ii)

02

Step 2. Write the Heron's formula.

We know, K=12ah,where,

h is the altitude to side a.

We also know, the Heron's formula is K=ss-as-bs-c

03

Step 3. Consider the given diagram.

Consider the given diagram,

鈻�PQRcan be separated into three smaller triangles 鈻�OPQ,鈻�OQR,鈻�OPR.

In 鈻�OPQ, side r is the altitude to side c.

In 鈻�OQR, side r is the altitude to side a.

In 鈻�OPR, side r is the altitude to side b.

Total area of 鈻�PQRis equal to the sum of these three triangles.

For 鈻�OQR, using equation (i),

KOQR=12ar

For 鈻�OPR, using equation (i),

KOPR=12br

For 鈻�OPQ, using equation (i),

KOPQ=12cr

04

Step 4. Add the three values of area.

Add the three values of area,

K=KOQR+KOPR+KOPQK=12ar+12br+12crK=12ra+b+c

From equation (ii),

K=rs

Hence, proved.

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