New SAT Math Workbook

# Wwwpetersonscom 226 chapter 13 exercise 5 1 a angle

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ions, the wheel will cover 20(3π) or 60π feet. 5. (D) Area of rectangle = b · h = 36 Area of square = s2 = 36 Therefore, s = 6 and perimeter = 24 5. (C) radius of circle = 3 Area = πr2 = 9π www.petersons.com Geometry 225 Exercise 3 1. (A) Exercise 4 1. (A) Find the midpoint of AB by averaging the x coordinates and averaging the y coordinates. 6 + 2 2 + 6 2 , 2 = ( 4, 4 ) 2. 14 x = 140 x = 10 (C) O is the midpoint of AB. x + 4 = 4, x = 0 y + 6 = 2, y = −4 The rectangle is 30′ by 40′. This is a 3, 4, 5 right triangle, so the diagonal is 50′. 2. (C) The altitude in an equilateral triangle is 1 always side ⋅ 3 . 2 x+4 =2 2 y+6 =1 2 A is the point (0, –4). 3. (A) d= 3. (D) This is an 8, 15, 17 triangle, making the missing side (3)17, or 51. (8 − 4 )2 + (6 - 3)2 = = 16 + 9 = 25 = 5 4 2 + 32 4. (D) Sketch the triangle and you will see it is a right triangle with legs of 4 and 3. 4. (A) The diagonal in a square is equal to the side times 2 . Therefore, the side is 6 and the perimeter is 24. (C) Area = 1 1 ⋅b ⋅h = ⋅4 ⋅3 = 6 2 2 5. 5. (A) Area of a circle = πr2 r=4 πr2 = 16π Triangle ABC is a 3, 4, 5 triangle with all sides multiplied by 5. Therefore CB = 20. Triangle ACD is an 8, 15, 17 triangle. Therefore CD= 8. CB – CD = DB = 12. The point (4, 4) lies at a distance of (4 − 0)2 + (4 − 0)2 = 32 units from (0, 0). All the other points lie 4 units from (0, 0). www.petersons.com 226 Chapter 13 Exercise 5 1. (A) Angle B = Angle C because of alternate interior angles. Then Angle C = Angle D for the same reason. Therefore, Angle D = 30°. (D) Exercise 6 1. (D) Represent the angles as x, 5x, and 6x. They must add to 180°. 12 x = 180 x = 15 2. The angles are 15°, 75°, and 90°. Thus, it is a right triangle. 2. (D) There are 130° left to be split evenly between the base angles (the base angles must be equal). Each one must be 65°. (E) 3. Extend AE to F. ∠A = ∠EFC ∠CEF must equal 100° because there are 180° in a triangle. ∠ AEC is supplement...
View Full Document

## This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.

Ask a homework question - tutors are online