Victory for the ACT Exam 16e TG Sample

518 • M ATH

STUDENT TEXT, p. 189 23. In the �igure below, if QRST is a square and the length of PQ is 20 , what is the length of RU ?

EXPLANATIONS

23. (D) Mathematics/Geometry/Complex Figures and Triangles/45°-45°-90° Triangles and 30°-60°-90° Triangles and Rectangles and Squares CC: HSG-MG.A.1 CCRS: ADV.G.3 Difficulty Level = 2; Teaching Time = 5–8 minutes; Purpose = Application Notice that the hypotenuse of PQT T is also a side of square QRST ; furthermore, RS it isias sniodteoonfly a side of the square, RSU T . If the length of QT is found, the length of RU can be deduced. Since PQT T is a 45°-45°-90° triangle and = PQ 2 0 , QT PQ 2 0 = = ^ h 2 2 20 = ^ h^ h . Since the four sides of a square are equal, 2 = = RS QT 0 . Since RS is the side in a 30°-60°-90° triangle ( ) RS TU = opposite the 30° angle, the length of RS is equal to RU 2 0 . Therefore, RS RU RU RU 2 2 2 4 & & = = = . 24. (H) Mathematics/Geometry/Complex Figures and Rectangles and Squares and Circles CC: HSG-MG.A.1 CCRS: EXP.G.2 Difficulty Level = 2; Teaching Time = 5–8 minutes; Purpose = Walkthrough What makes this item tricky is that it asks for the area of an odd-shaped �igure that dk eoye si sntoot sheaev tehaa tr et ha de ys-hma da de de fpoarrmt uo fl at .hTeh e �igure is what remains after the semicircle is subtracted from the square: To determine the shaded area, �irst �ind tNhoeteartehaast of the square and the semicircle. PS is not only a side of the sTqhuearreefo, irtei, ss itnhceedtihaemleetnegrtohfotfh e s e m i c i rc l e . PS equals 2, the area of the square is , s 2 4 2 2 = =

A. 20 B. 60 C. D. 4

2 2 0

24. In the �igure below, PQRS is a square, and PS is the diameter of a semicircle. If the length of PQ is 2, what is the area of the shaded portion of the diagram? F. 4 2 − r G. 4 0 − r

H. 4 2 - r J. 8 0 − r