   Chapter 6.CR, Problem 4CR ### 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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# Two circles intersect and have a common chord 12 cm long. The measure of the angles formed by the common chord and a radius of each circle to the points of intersection of the circles is 45 ° . Find the length of the radius of each circle.

To determine

To find: The radius length of each.

Explanation

Theorem:

1) A radius that is perpendicular to a chord bisects the chord.

2) If a line through the centre of a circle bisects a chord other than a diameter, then it is perpendicular to the chord.

Calculation:

Let the centers of the two circles be A and B.

And also let the common chord be PQ.

Given that the length of common chord = PQ = 12 cm.

The distance between the centers of the circles = AB.

Let the radius of the circle with center A = PA = x in.

Also, the measure of the angles formed by the common chord and a radius of each circle to the points of intersection of the circles is 45°.

So, APO=BPO=45°

But, APB=APO+BPO=45°+45°=90°

So, ◿PAB is a right triangle.

By theorem, OPAB.

So, ◿OAP and ◿OAB are also right triangles.

Since AOP=90°, PAO=PBO=45°

So, AO = PO and PO = OB

So, ◿OAP and ◿OAB are isosceles right triangles.

So, ◿OAP and ◿OAB are congruent by S

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