   Chapter 8.5, Problem 37E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# In a triangle with sides of lengths a, b and c and semiperimeter s, show that the length of the radius of the inscribed circle is r = 2 s s - a s - b ( s - c ) a + b + c

To determine

To prove:

The length of radius of circle inscribed in a triangle is r=2×s(s-a)(s-b)(s-c)a+b+c.

Explanation

Formula:

Area of a triangle with an inscribed circle:

If P is the perimeter of the triangle and r is the length of radius of its inscribed circle, then the area A of the triangle is given by

A=12rP

Area of triangle:

If a, b and c are lengths of sides of triangle, then area of triangle is given by the formula:

A=s(s-a)(s-b)(s-c)

Where s is the semi perimeter which is given by s=12(a+b+c)

Calculation:

To find the radius of inscribed circle, we will use the following formula:A=12rP, where A is the area of triangle, P is the perimeter of triangle and r is the radius of inscribed circle.

So, let’s find the area and perimeter of the triangle.

The lengths of sides of triangle is given as a, b and c

Perimeter of triangle is sum of lengths of all sides of triangle.

Thus, perimeter P=a+b+c.

Let A be the area of triangle.

The semi perimeter is half of the perimeter of triangle. Thus, s=a+b+c2.

Let’s substitute these values in the formula to find area of triangle.

A=s(s-a)(s-b)(s-c)

Let’s substitute the values of A and P in the formula: A=12rP

s(s-a)(s-b)(s-c)=12×r×(a+b+c)

Multiply both sides by 2: 2

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