SAT SUBJECT TEST MATH LEVEL 1
Quadrilaterals and Other Polygons
Questions 1 and 2 refer to the following figure, in which M and N are midpoints of two of the sides of square PQRS.
1. What is the perimeter of trapezoid PMNR?
(B) 2 + 3
(C) 3 +
2. What is the area of trapezoid PMNR?
3. The length of a rectangle is 5 more than the side of a square, and the width of the rectangle is 5 less than the side of the square. If the area of the square is 45, what is the area of the rectangle?
4. In rhombus PQRS, the ratio of mP to mQ is 1 to 5 and PQ = 6. What is the area of the rhombus?
5. If the length of a rectangle is 5 times its width and if its area is 180, what is its perimeter?
6. What is the average (arithmetic mean) of the measures of all the interior angles in a decagon?
7. If the interior angles of a pentagon are in the ratio of 2 : 3 : 3 : 5 : 5, what is the measure of the smallest angle?
8. In the figure below, the two diagonals divide square WXYZ into four small triangles. What is the sum of the perimeters of those four triangles?
(A) 4 + 4
(B) 16 + 8
(C) 16 + 16
9. How many sides does a polygon have if the measure of each interior angle is 9 times the measure of each exterior angle?
10. What is the area of trapezoid ABCD in the figure below?
11. What is the area of a regular hexagon whose sides are 4?
12. The area of a regular octagon whose sides are 2 can be expressed as a+b. What is the value of a + b?
Each of the problems in this set of exercises is typical of a question you could see on a Math 1 test. When you take the model tests in this book and, in particular, when you take the actual Math 1 test, if you get stuck on questions such as these, you do not have to leave them out—you can almost always answer them by using one or more of the strategies discussed in the “Tactics” chapter. The solutions given here do not depend on those strategies; they are the correct mathematical ones.
See Important Tactics for an explanation of the symbol ⇒, which is used in several answer explanations.
1. (B) Since M and N are midpoints of sides of length 2, PM = MQ = QN = NR = 1. Since is the hypotenuse of an isosceles right triangle whose legs are 1, MN = . Similarly, PR = 2, since it is the hypotenuse of an isosceles right triangle whose legs are 2. So the perimeter of trapezoidPMNR is 1+ +1+2=2+3.
2. (A) Even if you know the formula for the area of a trapezoid (and you should), the best way to proceed is to subtract the areas of the two white triangles from 4, the area of the square. The area of , and the area of Therefore, the area of the shaded region is 4 – 2 – 0.5 = 1.5.
3. (A) Let x represent the side of the square. Then the dimensions of the rectangle are (x + 5) and (x – 5), and its area is (x + 5)(x – 5) = x 2 – 25. Since the area of the square is 45, x 2 = 45, and so x 2 – 25 = 20.
4. (C) Draw and label the rhombus.
Since angles P and Q are supplementary:
m∠P + m∠Q = 180° ⇒ 5x + x = 180 ⇒ 6x = 180 ⇒ x = 30
Then since ΔPQT is a 30-60-90 right triangle, . Finally, the area of rhombus PQRS = bh = (PS)(QT) = (6)(3) = 18.
5. (D) Draw a diagram and label it.
Since the area is 180, we have . So the width is 6, the length is 5 6 = 30, and the perimeter is 2(6 + 30) = 2(36) = 72.
6. (E) By KEY FACT I2, the sum of the measures of the 10 interior angles in a decagon is (10 – 2) 180° = 8(180°) = 1,440°. So the average of their measures is 1,440° 10 = 144°.
7. (C) By KEY FACT I2, the sum of the angles of a pentagon is
(5 – 2) 180° = 3 180° = 540°
Let the degree measures of the five angles be 2x, 3x, 3x, 5x, and 5x. Then
The degree measure of the smallest angle is 2x, and 2 30° = 60°.
8. (C) Since the diagonals of a square are perpendicular, congruent, and bisect each other, each small triangle is a 45-45-90 right triangle whose hypotenuse is 4. Therefore, the legs are each . So the perimeter of each small triangle is 4 + 2 + 2 = 4 + 4 , and the sum of the perimeters is 4(4 +4) = 16 + 16.
9. (E) The sum of an interior and exterior angle is 180°. So 180 = 9x + x = 10x x = 18. Since the sum of all the exterior angles is 360, there are 360 18 = 20 exterior angles, 20 interior angles, and 20 sides.
10. (C) Draw in height AE. Then ΔAED is a 30-60-90 right triangle whose hypotenuse is 8. So DE = 4 and AE = 4. Then the area of trapezoid ABCD is .
11. (C) Since you do not know a formula for the area of a hexagon, you have to divide the hexagon into manageable pieces. There are many ways to do this. One way is to divide it into two trapezoids; another is to divide it into a rectangle and two triangles. The simplest way, though, is to divide it into six equilateral triangles.
By KEY FACT H11, the area of an equilateral triangle whose sides are 4 is . So the area of the hexagon is .
12. (C) Since you do not know a formula for the area of an octagon, you have to divide the octagon into manageable pieces. There are many ways to do that. One way is to draw four diagonals that divide the octagon into four isosceles right triangles, four rectangles, and a square.
The legs of the triangles are each . Therefore, each triangle has area 1, each rectangle has area 2, and the square has area 4. So the total area is 4(1) + 4(2) + 4 = 8 + 8. So a = 8, b = 8, and a + b = 16.