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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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Circle P has a radius of length 8 i n . Points A , B , C , and D lie on circle P in such a way that m A P B = 90 ° and m C P D = 60 ° . How much closer to point P is chord A B ¯ than chord C D ¯ ?

To determine

To find:

The distance that AB¯ is closer than CD¯ to point P.

Explanation

Given:

Circle P has a radius of length 8in. Points A, B, C, and D lie on circle P in such a way that mAPB=90° and mCPD=60°.

Theorem Used:

According to the Pythagorean theorem, in a right-angled triangle,

hypotenuse2 = base2 + perpendicular2.

Property Used:

If two sides of a triangle are equal then the angles formed by the legs of the included angle are also equal.

The sum of all the angles of a triangle is 180°.

An equilateral triangle has all the angles and all the sides equal.

The perpendicular bisector divides the chord into two equal parts.

Approach:

i) Find the length of the chords, AB¯ and CD¯.

ii) Use the Pythagorean theorem to calculate the distance of chords from center (i.e.) PM¯ and PN¯.

iii) Find the difference between, PM¯ and PN¯.

Calculation:

The central angles equals mAPB=90°, mCPD=60° and the radii equals PB¯=PC¯=8.

In the given figure,

Let M and N be perpendicular bisector on chords, AB¯ and CD¯ respectively.

Applying Pythagorean theorem, in APB since it is right angled,

hypotenuse2=base2+perpendicular2.

AB¯2=AP¯2+PB¯2 or AB¯=AP¯2+PB¯2

Substituting PB¯=AP¯=8 in AB¯=AP¯2+PB¯2,

AB¯=(8)2+(8)2=64+64=128=82

In CPD, the vertex angle is mCPD=60°, and sides PC¯=DP¯=8 are equal so the base angles are also equal.

Since, the sum of all the angles of a triangle is 180°. Hence, the sum of base angles is

mPDC+mPCD=180°60°=120°

Since, mPDC=mPCD,

mPDC=mPCD=120°2=60°

Thus, CPD is an equilateral triangle since all the angles are equal

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