   Chapter 8.2, Problem 17E ### 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 Exercises 13 to 18, find the area of the given polygon. kite A B C D with B D = 12 m ∠ B A C = 45 ∘ ,   m ∠ B C A = 30 ∘

To determine

To Find:

Area of the kite.

Explanation

Formula Used:

The area A of a kite whose diagonals have lengths d1 and d2 is given by A=12d1d2.

Calculation:

It is given that the kite ABCD has one of the diagonals BD=d1=12 and mBAC=45°,  mBCA=30°.

Let’s draw another diagonal AC so that the diagonals BD and AC meet at a point be 'O'.

AC=AO+OC  (1)

In kite, the diagonals d1 and d2 are perpendicular bisector of each other. That is BOA=BOC=90°.

As BD is the perpendicular bisector BO, BO=BD2=122=6

Now consider the triangle AOB,

ABO=180-(BOA+BAO)

ABO=180-(90°+45°)

ABO=45°

Therefore, the triangle AOB is a 45°:45°:90° right triangle

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