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# PLS HELP, DUE TMROW

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1)Chords WY and XZ of a circle are perpendicular. If XV = 4, WV = 3, and VZ = 9, then find YZ.

Picture: https://latex.artofproblemsolving.com/6/1/7/617805ac8c9a1eda9bbf996f32b9282efbd74701.png

2)Chord BA bisects chord OP at S. If AS = 3 and BS = 4, find OP.

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?

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.

Picture: https://latex.artofproblemsolving.com/b/f/3/bf3a913d7231e67c0610502487e0f4d2bbb4745f.png

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?

Picture: https://latex.artofproblemsolving.com/f/2/9/f2914a30663db28178f1400e8a0e4a809b6fd2e3.png

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

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

Picture: https://latex.artofproblemsolving.com/7/a/5/7a599f15c79349567f1b440d1e75d954dc5077a4.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.

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

Mar 6, 2018

#1
+1

Hi F

i can only help with 2 and 3. I can't see the diagrams of the others.

using the chord intercept theorem in both you get OP = sqrt48 and RX = 2

regards

Mar 7, 2018
#2
+100588
+2

1.  Intersecting  Chord Theorem :

WV  * VY   = XV * VZ

3 * VY  = 4 * 9

VY  =  4 * 9 /3     =  12

Since it appears the  VY  and VZ  are perpendicular....then....by the Pythagorean Theorem

VY^2  + VZ^2  =  YZ^2

12^2  +  9^2  = YZ^2

225  =  YZ^2   take the square root of both sides

15  = YZ

Mar 7, 2018
#3
+100588
+2

2)Chord BA bisects chord OP at S. If AS = 3 and BS = 4, find OP

O

B    S    A

P

Intersecting Chord Theorem

BS * AS  =   OS  * PS           but   OS  = PS....so

BS * AS  =  OS * OS

3 * 4  =  OS^2

12  = OS^2

√12  = OS   =  2√3

But  OP  = OS + PS = OS + OS  =    2OS    =   2 *  2√3     = 4√3

Mar 7, 2018
#4
+100588
+2

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

Mar 7, 2018
#5
+100588
+2

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

Therefore

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

Mar 7, 2018
#6
+100588
+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

So.....

PB + AB  = PA

9/5  +  AB  = 15

AB  =  15  - 9/5

AB  =   [ 75 - 9 ] / 5     =    66/5

Mar 7, 2018
#7
+100588
+2

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

Draw SC

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

Mar 8, 2018