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Elementary Geometry for College St...

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

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BuyFindarrow_forward

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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In Exercises 27 to 30, find the area of the figure shown.

Given: In O , O A = 5 ,
B C = 6 , a n d
C D = 4
Find: A A B C D

Chapter 8.2, Problem 27E, In Exercises 27 to 30, find the area of the figure shown. Given: In O, OA=5, BC=6, and CD=4 Find:

To determine

To Find:

The area of the quadrilateral ABCD inscribed by the circle with centre 'O'.

Explanation

Formula Used:

Angle subtended by a diameter on any point of circle is 90°.

Pythagorean theorem for the right angle triangle ABC for the hypotenuse AC,

AC2=AB2+BC2.

Area of right triangle =12×b×h

Calculation:

Consider the quadrilateral ABCD.

This quadrilateral can be divided into two triangles. That is, ABC and ADC.

By the angle subtended in a diameter of the circle theorem, it makes 90° at the mD and mB.

It is given that OA=5=radius, BC=6 and CD=4.

As radius is 5, the diameter is twice the radius.

CA=10

Consider the ABC,

Here BC=6 and CA=10. Applying Pythagorean theorem,

CA2=AB2+BC2

102=AB2+62

AB2=100-36

AB=8 units

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