S.A.T. Math Testing Tactics

S.A.T. Math Testing Tactics Tactic 9: Subtract To Find the Shaded Region

THINK OBJECTIVELY Many times you will need to find an area or perimeter but will not have all the information you need. These problems are testing you to see if you can use what you DO know to find what you need. One strategy to solve this type of problem is to subtract the unwanted portion of figure from the entire figure.

READ DIRECTIONS CAREFULLY! B • Find AREA of the white part of this rectangle (B) 1. Find the Area of the Entire rectangle 2. Subtract the area of Square A • Find PERIMETER of B 1. Find Perimeter of entire rectangle. 2. Subtract one side length of A. 3. Add in 3 sides lengths of A. A

Example 9.1 • In the figure, ABCD is a rectangle, and arc BE and CF are arcs of circles centered at A and D. • What is the area of the shaded region? Note: Figure not drawn to scale. A) 10 – π B) 2(5 – π) C) 2 (5 – 2π) D) 6 + 2π E) 5(2 – π)

Example 9.1 What is the area of the entire rectangle shape? A = L x W A = 5 x 2 = 10 What is the area of the white part (the unwanted part)? White part = Two ¼ circles (a total of one ½ circle) A = ½ π r2 A = ½ π (2)2 A = 2π Entire Shape – Unwanted Part = Shaded Region 10 – 2π 2(5 – π) Note: Figure not drawn to scale. A) 10 – π B) 2(5 – π) C) 2 (5 – 2π) D) 6 + 2π E) 5(2 – π)

Example 9.2 In the figure, each side of square ABCD is divided into three equal parts. If a point is chosen at random inside the square, what is the probability it will be in the white region? What are we looking for? Probability in white region = Area of white Region ÷ Area entire shape

Example 9.2 Visualize “pushing” the shaded regions together. They form a square of side x. Area of Shaded Region = x2 Area of Entire Region = (3x)(3x) = 9x2 White Region: 9x2 –x2 = 8x2 Probability of Landing in White Region = 8x2 ÷ 9x2 Or 8/9

JUST BECAUSE A PROBLEM HAS A SHADED REGION DOES NOT MEAN YOU MUST USE THIS TACTIC. If the previous problem had said “What is the probability of landing in the shaded region?” we wouldn’t have needed to do any subtracting. The probability of landing in the shaded region would be 1/9.

2 A B P O C D Example 9.3 Although this problem has no “shaded region,” We can still use the basic principle to solve the problem. First, try to find the length of OA. Second, find the length of OP. Finally, subtract: OA – OP = AP In the figure, the circle with center O is inscribed in square ABCD. Line segment AO intersects the circle at P. What is the length of AP? A) 1 B) 2 – √2 C) 1 – (√2/2) D) 2√2 – 2 E) √2– 1

2 A B P O C D Example 9.3 1 x 1 OP = Radius of circle Draw in another radius OX. If the side of the square is 2, Half the way across would be 1. So, the radius equals 1. By similar reasoning, AX = 1 Triangle AXO is an isosceles right triangle (45-45-90). If the side is 1, then the hypotenuse (OA) = √2 OX = 1 OP = 1 AX = 1 OA = √2 OA – OP = AP √2 – 1 = AP A) 1 B) 2 – √2 C) 1 – (√2/2) D) 2√2 – 2 E) √2– 1

In Conclusion If there is no direct path to finding the answer you need, try using an indirect approach. “Subtract to find the shaded region” is just a thought process for finding a solution to a problem where (at first) there doesn’t seem to be enough information to solve.

S.A.T. Math Testing Tactics