SAT Practice Questions - Geometry

Step 1: The edge of the paper will form the circumference of the base of
    the cylinder.
    
Step 2: Volume of cylinder = area of base × height
    
    Height = edge of the paper = 4
    

Answer: (B)

        
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

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°

Step 1: We can find the supplementary angle of 120°, which gives us 60°.

Step 2: is parallel to because they are both perpendicular to . This means that since they are corresponding angles.

Step 3: Triangle ABE and triangle ACF are similar triangles , since and (AA rule). Given that AE = EF, we can conclude that AB = BC

Step 4: Triangle ABE and triangle CBE are congruent triangles (SAS rule). So, x = angle AEB = 60°

So, x = 60

Answer: (E) 60

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 = πr². 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