SAT Mc Graw Hill 2011

CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 403

Concept Review 8

1. circumference = 2 πr


2. area =πr^2


3. A tangent line is a line that intersects a circle at
   only one point.


4. Any tangent to a circle is perpendicular to the
   radius drawn to the point of tangency.

Answer Key 8: Circles

3. 8 Draw a figure as shown, including the tangent
   segment and the radius extended to the point of
   tangency. You can find xwith the Pythagorean
   theorem:
   62 +x^2 = 102
   Simplify: 36 +x^2 = 100
   Subtract 36: x^2 = 64
   Take the square root: x= 8

SAT Practice 8

5. Draw BC

–––

to make a right triangle, and call the

length of the radius r.Then you can use the

Pythagorean theorem to find r:

82 +r^2 =(r+6)^2

FOIL: 64 +r^2 =r^2 + 12 r+ 36

Subtractr^2 :6 4 = 12 r+ 36

Subtract 36: 28 = 12 r

Divide by 12: 7/3=r

The circumference is 2πr,which is 2π(7/3)= 14 π/3.

A C

D

B

8

(^6) r
r

6. Draw the segments shown
   here. Since PNis a ra-
   dius of both circles, the
   radii of both circles have
   the same length. Notice
   thatPN, PR, RN, PT,and
   NTare all radii, so they are
   all the same length; thus, ∆PNT
   and∆PRN are equilateral triangles and their
   angles are all 60°. Now you can find the area of
   the left half of the shaded region.
   This is the area of the sector
   minus the area of ∆RNT.Since
   ∠RNT is 120°, the sector is
   120/360, or^1 ⁄ 3 , of the circle. The
   circle has area 36π, so the sec-
   tor has area 12π.∆RNTconsists
   of two 30°-60°-90°triangles, with
   sides as marked, so its area is
   . Therefore, half of the origi-

nal shaded region is , and the whole is

24 π−18 3.

12 π−9 3

()12 6 3 3 9 3/()()=

P N

R

T

S

N

R

T

6

6

3 √ 3

3 √ 3

3

A B BA A B

1. E The figure above demonstrates all three
   possibilities.


2. D The circumference = 2 πr= 16 π. Dividing by 2π
   givesr=8. Area =πr^2 =π(8)^2 = 64 π.

A

B

6

10

x