shapes are similar if they have the same angle mesurments,
similarity transformation says if you rotate a shape and dialate(change the size),
and it holds the same angles, then they are similar.
Obviously all circles are 360 degrees,
rotating a circle changes nothing, so all circles are similar.
s is some unknown amount bigger than r
which is the dialation.
So what they want, i assume, is that after a rotation and dilation, the angles of circle a and circle b,
remain the same and r and s are proportional to each other, as will all radius be,
so all circles are similar.
7) If AB = 3, AC = 8 and BS = 21, find the area of the circle.
AS * AB = AP * AC
(AB + BS) * AB = (AC + CP) * AC
24 * 3 = (8 + CP) * 8
72 = 64 + 8*CP
8 = 8*CP
PC = 1
AP = PC + AC
AP = 1 + 8 = 9
SP = √ ( AS^2 - AP^2) = √ ( 24^2 - 9^2 ) = √495
So..... SC = √ ( SP^2 + PC^2 ) =√ (495 + 1) = √ 496
And SPC is right....so SC is a diameter.of the circle.....so the radius is √496/2
So.....the area of the circle is
pi * (√496 / 2)^2 = 496 * pi / 4 units^2 = 124 pi units^2 ≈ 389.56 units^2
5)Point Y is on a circle and point P lies outside the circle such that PY is tangent to the circle. Point A is on the circle such that segment PA meets the circle again at point B. If PA = 15 and PY = 9, then what is AB?
This is known as the secant-tangent theorem.....specifically
PY^2 = AP * PB
9^2 = 15 * PB
81 = 15 * PB
PB = 81/15 = 9/ 5
PB + AB = PA
9/5 + AB = 15
AB = 15 - 9/5
AB = [ 75 - 9 ] / 5 = 66/5
Thank You SOOOO Much!!!! You have helped me with so many other questions had always respond so quickly. I really appreciate the time you take to help me and so many other people who are struggling with math and instead of just giving them an answer you go step by step in the problem to make sure they understand. Thank You So Much!!!!
First one :
A radius meeting a tangent does so at right angles.....so
angle ACB and angle ABC are complementary angles in right triangle ABC
24 + angle ABC = 90 subtract 24 from both sides
angle ABC = 66°
PS - if this is from VixenGrace....I answered your other multi-part question from yesterday.......
4)Let BC and DE be chords of a circle, which intersect at A, as shown. If AB = 3, BC = 15, and DE = 3, then find AE.
Draw BE and CD
We have two similar triangles
Angle BCD = angle BED and angle A is common to both
ΔCAD is similar to Δ EAB ⇒ CA / AD = EA / AB
Let AD = x and we have
18 / x = ( x + 3 ) / 3 cross-multiply
18 * 3 = x ( x + 3)
54 = x^2 + 3x rearrange
x^2 + 3x - 54 = 0 factor
(x + 9) ( x - 6 ) = 0
Setting the second factor = 0 and solving for x produces a positive result for x ⇒ 6 = AD
3)Chords PQ and RS of a circle meet at X inside the circle. If RS = 38, PX = 6, and QX = 12, then what is the smallest possible value of RX?
RX * XS = PX * QX
RX * XS = 6 * 12
RX * XS = 72
And RX + XS = 38 ⇒ XS = 38 - RX.....so....
RX * XS = 72
RX * (38 - RX ) = 72 rearrange
RX^2 - 38RX + 72 = 0 Factor
(RX - 36) (RX - 2) = 0
And its clear that the smallest value for RX = 2