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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 6.CR, Problem 4CR
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Elementary Geometry for College St...

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

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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, â—¿PAB is a right triangle.

By theorem, OP⊥AB.

So, â—¿OAP and â—¿OAB are also right triangles.

Since ∠AOP=90°, ∠PAO=∠PBO=45°

So, AO = PO and PO = OB

So, â—¿OAP and â—¿OAB are isosceles right triangles.

So, â—¿OAP and â—¿OAB are congruent by S

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Sect-6.1 P-1ESect-6.1 P-2ESect-6.1 P-3ESect-6.1 P-4ESect-6.1 P-5ESect-6.1 P-6ESect-6.1 P-7ESect-6.1 P-8ESect-6.1 P-9ESect-6.1 P-10E
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