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If AB = 3, AC = 8, and BS = 21, find the area of the circle.

https://imgur.com/SjicCE3

 

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.
https://imgur.com/F9koQXf

 

Let O be the center of a circle with diameter AB. Let C be a point on the circle such that angle COA = 90. Point P lies on line segment OA, and when line segment CP is extended past P, it intersects the circle at Q. If PQ = 7 and PO = 20, then find r^2, where r is the radius of the circle.

https://imgur.com/TmIyhb9

 Mar 5, 2024
 #1
avatar+128406 
+1

First one

 

BA (BA + BS)  = AC ( AC + PC)

3 ( 3 + 21)  = 8 ( 8 + PC)

72  = 8 (8 + PC)      divide both sides by 8

9  = 8 + PC

PC  = 1

 

Pythagorean Theorem

AS = 24

AP = 9

 

sqrt (AS^2 - AP^2)  = SP  =  sqrt [ 24^2 - 9^2]  = sqrt (495)

 

Pythagorean Theorem again

 

SC =  sqrt ( SP^2 + PC^2 )  = sqrt ( 495 + 1)  = sqrt (496)  =   4sqrt (31)

 

Because SPC is a right angle,  then  SC is the diameter of the circle

 

Then the radius is  2sqrt (31)

 

Area of circle  = pi * (2sqrt (31))^2  =  124 pi

 

cool cool cool

 Mar 5, 2024
 #2
avatar+2 
+1

BP = 22/3

 

r^2 = 624

 

Both solved with CAD, probably not how you're supposed to be doing it but I think it's right

 Mar 5, 2024
 #3
avatar+128406 
+1

Second one

 

There may be a way better way to  solve this !!!

 

Using the Law of Cosines extensively

 

-cos DCB   =   cos DAB

 

Draw DB

 

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

DB^2  = 6^2  + 2^2  + 2(6*2) cos (DCB)

 

100 + 9  - 60(cosDCB)  =  36 + 4 + 24cos(DCB)

69 = (24 + 60)cos (DCB)

69/84  = cos (DCB) = 23/28

sinDCB  = sqrt [ 28^2 - 23^2 ] / 28 = sqrt (255)/28 = sin DAB =  sin BCP

 

cosADC  = -cosABC

 

Draw AC

 

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

AC^2  =  2^2 + 3^2   + 2(2*3)cos ADC

 

100 + 36  - 120cosADC  = 4 + 9 + 12cosADC

136 - 120cosADC  = 13 + 12cos ADC

123  =  132 cos ADC

cos ADC = 123/132  = 41/44 

sin ADC =  sqrt (44^2 - 41^2) / 44 =  sqrt (255) / 44  = sin ABC = sin PBC

 

sin PBC / sin BCP  =   PC / BP =  [ sqrt (255)/ 44 ] / [ sqrt (255) / 28]

PC/BP  = 28/44  = 7/11

PC = (7/11)BP

 

BP^2  = BC^2  + PC^2  - 2(BC * PC)cosBCP

BP^2  = 3^2 + [ (7/11)BP]^2  - 2 [ 3 * (7/11)BP ] cos DAB

BP^2  = 9 + [ (7/11) BP ]^2  - (42/11)BP (-cosDCB)

BP^2 = 9 + [ 49/121]BP^2  + (42/11)(23/28) BP

 

Let BP  =  x

 

x^2 - (49/121)x^2 - 69/22x  - 9 =  0

 

(72/121)x^2  - (69/22)x - 9  =  0

 

Solving this  for x  produces  x  = 22 / 3   =  BP

 

 

cool cool cool

 Mar 6, 2024
 #4
avatar+128406 
+1

Last one

CO  = r

PO = 20

PC  =sqrt [ 20^2 + r^2 ]

AP  = AB  - PB  =  2r - [ r + 20 ] =  r - 20

PQ = 7

 

Product of intersecting chord segments

 

PQ * PC  = AP * PB

 

7 * sqrt (20^2 + r^2)  =  (r-20)(r + 20)

 

7* sqrt [ 400 + r^2 ]  = r^2 - 400           square both side

 

49 (400 + r^2)  =  r^4 - 800r^2 + 160000

 

19600 +  49r^2  = r^4 - 800r^2 + 160000

 

r^4  - 849r^2  + 140400  =  0

 

This produces two positive solutions for r

 

r  =15   (reject)  PO is already greater than 15

 

r = 4sqrt (39)   

 

r^2  =  16 * 39   =    624

 

cool cool cool

 Mar 6, 2024
edited by CPhill  Mar 6, 2024

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