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avatar+1442 

1:  Let line BC and line DE be chords of a circle, which intersect at A, as shown. If AB = 3, BC = 15, and DE = 3, then find AE.

 

2:  In the adjoining figure, line AB is tangent at A to the circle with center O, point D is inside the circle, and line DB intersects the circle at C. If BC = DC = 3, OD = 2, and AB = 6, then find the radius of the circle.

 

3:  In triangle ABC, AB = 10, AC = 8, and BC = 6. Let P be the point on the circumcircle of triangle ABC so that angle PCA = 45 degrees. Find CP.

 

4:  Let P be a point outside of circle O. A segment is drawn from P such that it is tangent to circle O at point T. Meanwhile, a secant from P intersects O at A and B, such that PA < PB. If PA = 3 and PT = AB - PA, then what is PB?

 

Thanks so much!

 #1
avatar+98061 
+1

1.  We can find AE  with the secant product theorem

 

CA * AB   =  AE * DE

 

18 * 3   =  AE * AD

 

54  =  AE * AD

 

Let  AD  =  x

Then AE  =  3 + x        and we have that

 

54  = (3 + x)* x

 

54  = 3x + x^2       rearrange as

 

x^2 + 3x - 54   =  0        factor

 

(x + 9) (x - 6)  = 0

 

Setting each factor to 0 and solving for x and we have that

 

x  = -9    (reject)          and      x  = 6  (accept)

 

So

 

x = AD  =  6

 

And AE  =  3 + x   =   3 + 6   =  9

 

 

 

cool cool cool

 May 3, 2018
 #2
avatar+98061 
+1

3.   Note that  triangle ABC is a 6-8-10 right triangle

And angle BCA  =  90

And angle PCA  =  45

So....angle BCP  = 45

Then.....PC  bisects  angle BCA

Let the intersection of PC and AB  =  D

So.....in triangle ABC, since angle BCA  is bisected, we have the following relationship

BC/CA   = BD/AD

6/8  = BD/AD

3/4  = BD/AD

 

So....since BA  = 10.....there  are 7 equal parts of BA

 

BD  =  (3/7)*10   =  30/7

AD = (4/7)* 10  = 40/7

 

Note....sin BAC  =  6/10  =  3/5

And  sin PCA = sin 45  =  1/√2

 

So.....using the Law of Sines

 

AD/ sin PCA  = CD/ sin BAC

 

(40/7)/ (1/√2)  =  CD / (3/5)

 

CD  =  (40/7)(3/5)*√2   =  (120/35)√2  =  (24/7)√2

 

And using the intersecting chord theorem

BD * AD  = CD * PD

(30/7) (40/7)  =  24√2 / 7  * PD

1200/7  =  24√2 * PD

PD  =  (1200) / ( 7 * 24 * √2) 

PD  =  (25/7)√2

 

So....CP  = 

PD + CD  =

(25/7)√2  + (24/7)√2  =

(49/7)√2 =

7√2

 

 

cool cool cool

 May 3, 2018
 #3
avatar+98061 
+1

4.  We can use the secant-tangent theorem to solve this, ACG...

 

We have something like this  (not to scale) :

 

 

PA * PB  = PT^2

 

PA * PB  = (AB - PA)^2

 

PA * ( AB + PA)  = AB^2 - 2PA*AB + PA^2

 

PA*AB + PA^2 = AB^2  - 2PA*AB + PA^2

 

3*AB + 3^2  = AB^2  - 2*3*AB  + 3^2

 

3AB + 9  = AB^2 - 6AB + 9

 

3AB  = AB^2 - 6AB     rearrange as

 

AB^2  - 9AB  = 0       factor

 

AB( AB - 9)  =  0

 

Seting each factor to 0 and solving for AB  we have that either

 

AB  = 0  (reject)    or  AB  = 9   (accept)

 

So.....PB  = AB + PA

 

PB  = 9 + 3

 

PB  = 12

 

 

cool cool cool

 May 5, 2018
 #4
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0

CPhill: Please check my reasoning and see where I went wrong. Thanks.

3)

Triangle ABC is a right-angle triangle. Angle BCA is 90 degrees. This angle is bisected from C to  a point D on AB. The height of this line CD is 4.8. In triangle ABC, angle BCA =36.87, angle ABC=53.13.

 

In Triangle ACD, AD =8 / sin(45) = 5.657 and BD = 10 - 5.657 =4.343. But we have:

5.657 x 4.343 =4.8 x PD, and PD =5.11845. AD + PD =4.8 + 5.11845=9.91845 =~7Sqrt(2).

Note. But the answer isn't exactly 7sqrt(2) !!. Why? Can you spot the mistake?.

 May 5, 2018
edited by Guest  May 5, 2018

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