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6)Two circles intersect at A and B. A common external tangent is tangent to the circles at T and U as shown. Let M be the intersection of line AB and TU. If AB = 9 and BM = 3 find TU.

Picture: https://latex.artofproblemsolving.com/c/5/0/c50a548001dc09c6eb63031398f6a77787276a50.png

 

8)Circle O is tangent to AB at A, and angle ABD = 90 degrees. If AB = 12 and CD = 18, find the radius of the circle.

Picture: https://latex.artofproblemsolving.com/f/f/1/ff13ec3e917b2a46649871ef2354fc1290478bc0.png

 

9)In cyclic quadrilateral ABCD, AB = 2, BC = 3, CD = 10, and DA = 6. Let P be the intersection of lines AB and CD. Find the length BP.

Picture: https://latex.artofproblemsolving.com/d/a/2/da27e9a157dbaae3feaf8dce05d22b710c368627.png

 

10)Let A, B, C be points on circle O such that AB is a diameter, and CO is perpendicular to AB. Let P be a point on OA, and let line CP intersect the circle again at Q. If OP = 20 and PQ = 7, find r^2, where r is the radius of the circle.

Picture: https://latex.artofproblemsolving.com/5/2/0/52028bb1d6c8a816ada5d2be3e55df12d704da4a.png

 

11)Two parallel chords in a circle have lengths 18 and 14, and the distance between them is 8. Find the length of the chord parallel to these chords and midway between them. (The diagram is not drawn to scale.)https://latex.artofproblemsolving.com/c/0/6/c06a5e099c8cb008ed2811aa67bc29aa48e94b8a.png

FiestyGeco  Mar 8, 2018
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2+0 Answers

 #1
avatar+86604 
+1

8)Circle O is tangent to AB at A, and angle ABD = 90 degrees. If AB = 12 and CD = 18, find the radius of the circle.

 

We have the secant-tangent theorem, first

 

AB^2  = CB (CB + CD)

12^2  = CB^2 + 18CB

CB^2  + 18CB - 144   = 0    factor as 

 

(CB - 6)  ( CB + 18)  = 0

 

The frist answer  gives the positive answer  for  CB - 6 =  0  ⇒  CB  = 6

 

Draw AC    ......and the tangent of  angle BAC  =   6/12   =  1/2   

 

So   arctan (1/2)  = BAC....and minor arc  AC  is twice this  = 2 arctan (1/2) 

 

And  AC  =  √   ( BA^2  + BC^2)  = √  (12^2  + 6^2)  = √ (144 + 36)  =  √160

 

Draw  OC  and OA    and angle  COA  = minor arc AC  =  2 arctan (1/2) 

 

Using the Law of Cosines to find the radius.....we have that

 

160  = 2r^2  - 2r^2 * cos (2 arctan (1/2) )  

 

Let   cos (2arctan(1/2) )  =   1  - 2sin^2 ( arctan(1/2) ) ⇒  sin (arctan (1/2) )  = 1/ √5  ....so we have

 

  1  -  2(1/√5)^2   =  1 - 2/5  = 3/5  = .6 

 

160  = 2r^2 - 2r^2 (.6)

 

160  = r^2  ( 2 - 1.2)

 

160 = r^2 (.8)

 

200  = r^2

 

r  = √200  =    10√2

 

 

cool cool cool

CPhill  Mar 8, 2018
 #2
avatar+86604 
+1

9)In cyclic quadrilateral ABCD, AB = 2, BC = 3, CD = 10, and DA = 6. Let P be the intersection of lines AB and CD. Find the length BP.

 

Draw AC

 

AC^2  =  2^2  + 3^2   -  2(3)(2)cos (ABC)  

AC^2  = 13  - 12cos(ABC)   (1)

And since ABC is obtuse and ADC  is supplemental to ABC, cos(ADC)  = -cos(ABC)

So

AC^2   = 6^2  + 10^2 - 2(6)(10)cos(ADC)

AC^2  = 6^2 + 10^2  - 120 [ -cos(ABC) ]

AC^2  = 6^2 + 10^2 + 120cos(ABC)   

AC^2  = 136 + 120cos(ABC)   (2)

Subtract (1) from (2)

0  = 123  + (120 + 12) cos(ABC)

0  =  123  + 132cos(ABC)

-123 / 132  =  cos (ABC)

-41/44 = cos(ABC)   

So  sin (ABC)  =  √ [ 1 -  (41/44)^2 ]  =   √ (255) / 44   =  sin PBC

 

Likewise.....draw  DB

DB^2  = 6^2  + 2^2  - 2(6)(2)cos(DAB)

DB^2  = 40 - 24cos(DAB)

And sincw DAB is obtuse and DCB is supplemental to DCB, cos(DCB) = -cos(DAB)

So

DB^2  = 10^2 + 3^2  - 2(10)(3)cos(DCB)

DB^2 =  10^2 + 3^2  - 60 [-cos(DAB) ]  

DB^2   =  109 + 60cos(DAB)

Subtract (3) from (4)

0  = 69 + (60 + 24) cos(DAB)

-69/84  = cos(DAB)

-23/28  = cos(DAB)

So sin (DAB)  = √  [ 1 - (23/28)^2 ]  = √ (255)/28  =  sin PCB

 

sin PBC / sin PCB   = PC/ PB

(√ (255) / 44)  / ( √ (255)/28)  = PC / PB

28 / 44 = PC /PB

7/11 = PC /PB

PC   =  (7/11)PB

 

And we have that

 

(PB + AB) * PB   = PC(PC + CD)

(PB + 2) * PB  = (7/11)PB [ (7/11)PB + 10)

PB^2 + 2PB  =  (49/121)PB^2 + (70/11)PB

 

Let PB  = x

 

x^2 + 2x =  (49/121)x^2 + (70/11)x

(121 - 49) x^2 / 121   +  (2 - 70/11)x  = 0

(72/121)x^2  -  (48/11)x  = 0

(1/11)x [ (72/11)x - 48 ] = 0

 

So..... either x  =  0   {reject}    or

 

(72/11)x  - 48  = 0

(72/11)x  =  48

  x =  48(11)/72  =  (2/3) * 11  =   22/3   =  PB  = BP

 

Proof   {PC  = (7/11)PB  =  (7/11)(22/3)  = 14/3 }

 

(PB + AB) * PB   = PC(PC + CD)

 

(22/3 + 2) * (22/3)  =  (14/3) * ( 14/3 + 10)

 

616 / 9    =   616 / 9

 

 

 

 

cool cool cool

CPhill  Mar 9, 2018

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