+25 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
+128579 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
+2 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
+128579 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
+128579 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
