Step 1: Area of shaded region = Area of circle – area of square.
We need to get the area of the circle and area of the square. Step 2: The diagonal BD makes two 45°-45°-90° triangles with the sides of
the square. Step 3: Using the 45°-45°-90° special triangle ratio . If the leg is 2
then the diagonal BD must be . Step 4:BD is also the diameter of the circle.

Step 5: Area of square = 2 × 2 = 4 Step 6: Area of shaded region = Area of circle – area of square = 2π – 4

Answer: (B) 2π – 4

SAT Practice Questions - Geometry: 1,
2, 3,
4, 5,
6, 7, 8, 9, 10

Step 1:x and y are supplementary angles. So if x < 90° then y > 90° Beware! Although the figure seems to show that l is parallel to m, it is not
stated in the question.

Answer: (E) y > 90°

SAT Practice Questions - Geometry: 1,
2, 3,
4, 5,
6, 7, 8, 9, 10

Step 1: Arc AB = 40°. Circumference of circle = 360°
Let C = circumference of circle

Step 2: You are given that arc AB = 4π. Plug into above equation.

Step 3: Using the formula for the circumference of circle: C = 2πr. Plug into the above equation.
2πr = 36π
2r = 36 r = 18 Step 4: Using the formula for the area of circle: A = πr^{2}. Plug in value for r. A = π(18)^{2} = 324π Step 5: Sector AOB is of the area of the circle.
Sector AOB = × 324π = 36π

Answer: (C) 36π

SAT Practice Questions - Geometry: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Which segment is congruent to BE?

Which line has a slope of -3?

AD is parallel to BC. AC = 13, BC = 5, BD = 15. What is the length of AD?