Victory for the ACT Exam 16e TG Sample

U NIT 4 | G EOMETRY • 519

STUDENT TEXT, p. 189

EXPLANATIONS

and the semicircle’s radius is d 2 2 2 1 = = The area of a circle with radius 1 is (1) r = = r r r 2 2 , so the semicircle’s area is = 2 area circle 2 0 r . Finally, the area of the st hhea daer edapoofrtthi oensiqsutahree dainf fde rtehnecaer be ae towf et he ne semicircle: 4 − 2 r . 25. (D) Mathematics/Geometry/Complex Figures and Triangles/Properties of Triangles CC: HSG-MG.A.1 CCRS: EXP.G.2 Difficulty Level = 2; Teaching Time = 5–8 minutes; Purpose = Application Tt rhi aenl ge lneg; tdhisv iodfe3t, h4e, aqnuda d5r si lua gt egreaslt i an troi gthwt o triangles: .

25. If the lengths of the sides, in inches, are as marked on the �igure below, what iqsutahderailraetae,rainl?square inches, of the

A. B. 6 6 + 3 C. 12 D. 18

Ob ansee tor ifa3n, gsloe iht ahsaasna na l tairteuad eo fo f 4 a n d a . ( ) 2 3 4 6 = Talhtietuodtheeirs tnreiaendgelde, hsoasaaddbaits:e of 6, but an

Ba let ictauudsee dt hi vei dt reisa tnhgel ebiassies oi ns cheal el fs, ,ctrheea t i n g tl ewnog t3h- 44-. 5T thr ei ar ne fgol er es , wt hi tehs ae csohna dr etdr i laengg ol ef has an area of 4( ) 12 2 6 = . The area of the qtwuaodtrriilaantegrleasl:is the sum of the areas of the 6 12 18 .