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Chapter 9: Q15. (page 336)

PA炉, PB炉,and RS炉 are tangents.

Explain, Why PR+RS+SP=PA+PB.

Short Answer

Expert verified

PR+RS+SP=PA+PB

Step by step solution

01

Step 1. Given information.

The figure here given is,

PA炉,PB炉andRS炉are tangents

02

Step 2. Concept Used.

Theorem 9.1 Corollary,

Tangents to a circle from a point are congruent.

03

Step 3. Consider the given figure for further solution.

From the above figure,

it is clear that, RA炉and RC炉 are tangents to the circle from the common point R,so by theorem 9.1 corollary, it can be said that,

RA=RC 鈥�.. (1)

Similarly, BS炉and CS炉 are tangents to the circle from the common point \[S\],so by theorem 9.1corollary, it can be said that,

BS=CS 鈥�.. (2)

Now consider,

PA+PB=PR+RA+BS+SP

Then using 1 and 2,

PA+PB=PR+RC+CS+SP=PR+RC+CS+SP=PR+RS+SP

Therefore, it is proved that, PR+RS+SP=PA+PB.

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Most popular questions from this chapter

Draw a diagram similar to the one shown, but much larger. Carefully draw the perpendicular bisectors of AB炉 and BC炉.

  1. The perpendicular bisect in a point. Where does that point to be?
  2. Write an argument that justifies your answer to part (a).

Draw a circle with center O and a line TS鈫�tangent to at 鈯�O. Draw OT炉and use a protractor to find m鈭�OTS?

Plane Z passes through the center of sphere Q.

  1. Explain why QR=QS=QT.
  2. Explain why the intersection of the plane and the sphere is a circle. (The intersection of a sphere with any plane passing through the center of the sphere is called a great circle of the sphere).

For each exercise draw 鈯�O with radius 12. Then draw radii OA炉and to OB炉 form an angle with the measure named. Find the length of AB炉.

b. m鈭�AOB=180掳

Is there a theorem about spheres related to the theorem in Exercise 12? If so, state the theorem.
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