Chapter 8.2, Problem 56E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# For the cyclic quadrilateral MNPQ, the sides have lengths a, b, c, and d. If a 2 + b 2 =   c 2 + d 2 , explain why the area of the quadrilateral is A =   a b + c d 2 .

To determine

To Show:

The area of the quadrilateral MNPQ is A= ab+cd2 if it is a cyclic quadrilateral with sides a, b, c and d which satisfies the condition a2+b2= c2+d2.

Explanation

Formula Used:

1. Pythagorean theorem for the right angle triangle âˆ†ABC for the hypotenuse AC,

AC2=AB2+BC2.

2. Area of right triangle =12bh, where b is the base and h is the height of the triangle.

It is given that the cyclic quadrilateral MNPQ with sides a,Â b,Â c and d which satisfies the condition a2+b2=Â c2+d2.

Letâ€™s draw a diagonal between M and P, and MP = k.

Here we can split the quadrilateral MNPQ into two triangles MNP and MQP.

Since the quadrilateral MNPQ is cyclic and satisfies the condition a2+b2=Â c2+d2, its diagonal MP must be a diameter of the circle.

As MP is a diameter the triangles MNP and MQP are right triangles

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