   Chapter 8.5, Problem 26E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

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

For concentric circles with radii of lengths 3 in. and 6 in., find the exact area of the smaller segment determined by a chord of the larger circle that is also a tangent of the smaller circle.

To determine

To find:

The area of segment formed by chord of larger circle which is also a tangent of smaller circle.

Explanation

Formula:

Area of segment:

Asegment=Asector-A.

Area of sector:

If r is the radius of the circle, the area A of a sector whose arc has degree measure m is given by

A=m360×πr2.

Area of triangle:

If b and h are lengths of base and height of triangle respectively, then area A of triangle is given by the formula:

A=12bh.

Sine ratio:

In a right angled triangle, sine of an angle θ is given by:

sinθ=opp(θ)hypotenuse

Where opp(θ) means the length of side opposite to θ.

Calculation:

Consider the following diagram in which the radius of inner circle is OC. Hence, OC=3 in.

The radius of outer circle is OA and OB. Hence, OA=OB=6 in.

We have to find the area of the segment formed by AB- and AB^.

For finding the area of segment, first we need to find the area of sector formed by radii OA, OB and AB^

To find the area of segment, first we have to find the degree measure of AB^ which is equal to mAOB.

Let’s consider right angled AOC. In this triangle, sinOAC=OCOA=36=12

We know that sin30°=12

Hence, OAC=30°

Sum of all angles of a triangle is 180°

In AOC, OAC+ACO+AOC=180°

After substituting the values of ACO and OAC, we get AOC=60°

Consider OCA and OCB.

In these two triangles, OA=OB=6

Hence, OCA=OCB=90°

OC is common side for both the triangles.

From the above observations, it is clear that OCAOCB by RHS (right-angle hypotenuse side) test of congruency.

If two triangles are congruent, then corresponding angles are equal.

Hence, ∠AOC=BOC

Thus, BOC=60°

AOB=AOC+BOC=60+60=120°

The degree measure of AB^=mAOB=120°

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