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 $K = r s$ , where, $s = \frac{1}{2} (a + b + c)$ .
Then show that $r = \sqrt{\frac{(s - a) (s - b) (s - c)}{s}}$ .

Short Answer

Expert verified

To show that $r = \sqrt{\frac{(s - a) (s - b) (s - c)}{s}}$ , first find the area of the three triangles and add them, followed by simplification.

Step by step solution

Step 1. Given information.

Consider the given question,

$r = \sqrt{\frac{(s - a) (s - b) (s - c)}{s}} . . . . . . (i) s = \frac{1}{2} (a + b + c) . . . . . . (ii)$

Step 2. Write the Heron's formula.

We know, $K = \frac{1}{2} a h$ ,where,

h is the altitude to side a.

We also know, the Heron's formula is $K = \sqrt{s (s - a) (s - b) (s - c)}$

Step 3. Consider the given diagram.

Consider the given diagram,

$△ P Q R$ can be separated into three smaller triangles $△ O P Q, △ O Q R, △ O P R$ .

In $△ O P Q$ , side r is the altitude to side c.

In $△ O Q R$ , side r is the altitude to side a.

In $△ O P R$ , side r is the altitude to side b.

Total area of $△ P Q R$ is equal to the sum of these three triangles.

For $△ O Q R$ , using equation (i),

$K_{O Q R} = \frac{1}{2} a r$

For $△ O P R$ , using equation (i),

$K_{O P R} = \frac{1}{2} b r$

For $△ O P Q$ , using equation (i),

$K_{O P Q} = \frac{1}{2} c r$

Step 4. Add the three values of area.

Add the three values of area,

$K = K_{O Q R} + K_{O P R} + K_{O P Q} K = \frac{1}{2} a r + \frac{1}{2} b r + \frac{1}{2} c r K = \frac{1}{2} r (a + b + c)$

From equation (ii),

$K = r s$

Hence, proved.

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Short Answer

Step by step solution

Step 1. Given information.

Step 2. Write the Heron's formula.

Step 3. Consider the given diagram.

Step 4. Add the three values of area.

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