- E Use the formula: central angle/360 = arc length/circumference. Plug in what
you are told: central angle 360 = circumference3π. To solve for the central angle, you must
solve for the circumference.
area = πr^2 = 16π
r^2 = 16
r = 4
The diameter = 2r = 8. The circumference = 8π. Substitute 8π for the
circumference in the preceding proportion:
central angle 360 = 3π8π
central angle 360 =^38
central angle = (^38 )360 = 135 - C Since the central angle is 120 degrees, the arc length =^120360 × 2πx, and the
sector area =^120360 × πx^2. Thus
(^) area of sector AOBlength of arc AB =
120
360 × 2πx
120
360 × πx^2
=^13 × 1 2πx
3 × πx^2
= 2ππxx 2
=^2 xx 2
=^2 x
- C To determine the area of the shaded region, calculate the area of sector AOB
and subtract the area of triangle AOB. Since the central angle of sector AOB
is 90, the area of sector AOB = 36090 × πr^2 = 36090 × π6^2 =^14 (36π) = 9π. The area of
right triangle AOB =^12 × leg 1 × leg 2 =^12 × 6 × 6 = 18. The area of the shaded
region is thus 9π – 18. - 4 Since AO and AB are radii of the circle, they must be equal. Thus,
corresponding angles BAO and ABO must be equal. Since angle ABO = 60,
angle BO = 60. Thus angle AOB + 60 + 60 = 180 → angle AOB = 60. All three
angles of the triangle are equal, which means all three sides are equal. Thus the
length of AB is 4. - D When a right angle is inscribed in a semicircle, its hypotenuse = the
diameter of the circle. Thus AB = 8. Since ABC is a 45-45-90 triangle, each
of the legs = √2^8. The area of triangle ABC is thus^12 × (√2^8 )^2 =^12 ×^642 =
1
2 × 32 = 16.
Quantitative Comparison Questions
- B Since x < 90, the sector area will be less than 36090 × the area of the circle.
Thus the sector area will be less than 25% of the area of the circle. - C Since point Q is on the circle, the circle’s radius is equivalent to OQ. Note
that OQ is the hypotenuse of right triangle OPQ. Since the legs of the right
420 PART 4 ■ MATH REVIEW04-GRE-Test-2018_313-462.indd 420 12/05/17 12:05 pm