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OP. By Principle 1, j APO is a rt. j .
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OQ; hence, OPAQ is a rhombus with a right angle, or a square.
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(d) By Principle 1, AP ' PR and BR ' PR. Then AP i BR, since both are ' to PR. By Principle 1, AB ' OQ; also, PR ' OQ (Given). Then AB i PR, since both are ' to OQ. Hence, PABR is a parallelogram with a right angle, or a rectangle.
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6.6 Applying principle 1 (a) In Fig. 6-26(a), AP is a tangent. Find j A if mj A: mj O
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2:3. 140 .
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(b) In Fig. 6-26(b), AP and AQ are tangents. Find mj 1 if mj O
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(c) In Fig. 6-26(c), DP and CQ are tangents. Find mj 2 and mj 3 if j OPD is trisected and PQ is a diameter.
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(a) By Principle 1, mj P 90 . Then mj A x 18. Hence, m j A 36 . mj O 90 . If mj A 2x and mj O 3x, then 5x 90 and
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CHAPTER 6 Circles
Fig. 6-26
(b) By Principle 1, mj P mj Q 90 . Since mj P mj Q mj O mj A Since mj O 140 , mj A 40 . By Principle 5, mj 1 1 mj A 20 . 2 (c) By Principle 1, m j DPQ mj PQC of ^ PBQ, mj 3 90 60 150 . 90 . Since mj 1 30 , mj 2
360 , mj A
mj O
60 . Since j 3 is an exterior angle
6.7 Applying principle 4 (a) AP, BQ, and AB in Fig. 6-27(a) are tangents. Find y. (b) ^ ABC in Fig. 6-27(b) is circumscribed. Find x. (c) Quadrilateral ABCD in Fig. 6-27(c) is circumscribed. Find x.
Fig. 6-27
Solutions
(a) By Principle 4, AR (b) By Principle 4, PC 11 8 19. (c) By Principle 4, AS 10 8 18. 6, and RB 8, QB 10, CR y. Then RB 4, and AP 5, and RD AB AR AQ. Then AQ SD. Then RD 14 6 8. Hence, y RB 8. AP AS PC SD
AB QB CD CR
11. Hence, x 8. Hence, x
6.8 Finding the line of centers Two circles have radii of 9 and 4, respectively. Find the length of their line of centers (a) if the circles are tangent externally, (b) if the circles are tangent internally, (c) if the circles are concentric, (d) if the circles are 5 units apart. (See Fig. 6-28.)
Fig. 6-28
CHAPTER 6 Circles
Solutions
Let R radius of larger circle, r 9 and r 9 and r 4, OO9 4, OO9 radius of smaller circle. R R r r 9 9 4 4 13. 5.
(a) Since R (b) Since R
(c) Since the circles have the same center, their line of centers has zero length. (d) Since R 9, r 4, and d 5, OO9 R d r 9 5 4 18.
6.9 Proving a tangent problem stated in words
Prove: Tangents to a circle from an outside point are congruent (Principle 4). Given: Circle O AP is tangent at P. AQ is tangent at Q. To Prove: AP > AQ Plan: Draw OP, OQ, and OA and prove ^ AOP > ^ AOQ.
PROOF:
Statements 1. Draw OP, OQ, and OA. 2. 3. 4. 5. 6. OP > OQ j P and j Q are right angles. OA > OA ^ AOP > ^AOQ AP > AQ
Reasons 1. A straight line may be drawn between any two points. 2. Radii of a circle are congruent. 3. A tangent is to radius drawn to point of contact. 4. Reflexive property 5. Hy-leg s 6. Corresponding parts of congruent ^ are congruent.
6.3 Measurement of Angles and Arcs in a Circle
A central angle has the same number of degrees as the arc it intercepts. Thus, as shown in Fig. 6-29, a central angle which is a right angle intercepts a 90 arc; a 40 central angle intercepts a 40 arc, and a central angle which is a straight angle intercepts a semicircle of 180 . Since the numerical measures in degrees of both the central angle and its intercepted arc are the same, we may restate the above principle as follows: A central angle is measured by its intercepted arc. The symbol may be used to mean is measured by. (Do not say that the central angle equals its intercepted arc. An angle cannot equal an arc.)
Fig. 6-29
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