   Chapter 8.5, Problem 36E ### 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 right triangle with sides of lengths a, b and c (where c is the length of the hypotenuse), show that the length of the radius of the inscribed circle is r = a b a + b + c

To determine

To prove:

The length of radius of circle inscribed in a right angled triangle is r=aba+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).

Pythagoras theorem:

In a right angled triangle, if the length of hypotenuse is c and the length of remaining two sides of triangle is a and b, then by Pythagoras theorem, c2=a2+b2.

Expansion of a+b(a-b):

After expansion, the above expression can be written as a2-b2.

Expansion of a+b2:

a+b2=a+ba+b=a2+2ab+b2.

Expansion of a-b2:

a-b2=a-ba-b=a2-2ab+b2.

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 right angled 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)

A=a+b+c2a+b+c2-aa+b+c2-ba+b+c2-c

Simplify: A=a+b+c2b+c-a2a+c-b2a+b-c2

Simplify: A=14a+b+cb+c-aa+c-b(a+b-c)

Rearrange the products to simplify: A=14a+b+ca+b-cc-a-b[c+(a-b)]

Using the expansion formula, a+b-c(a+b-c) can be written as: a+b2-c2

a+b2-c2 can be further simplified as a2+2ab+b2-c2

It is given that c is the hypotenuse of the right angled triangle

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