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he area of the shaded region of the figure. Use 3.14 for π as necessary. 4) T 10 in. 30 in. 4= 1/2 bh ↓ A=TT(192) V ½/2 (30) (10)78.5 150 4) 15 in. ㅓ 1.5 A) 186 in.2 B) 146 in.2 C) 221 in.2 D) 71 in.2 Find the area. Use 3.14 for л as necessary. 5) Lookover 7 cm A=1/2bh -1/2(7.21) 13 cm (73.5) A = 1/2 (πr²) 12 (3.14-132) ↓ 265.33 265.33-735 = 191.83 ?!! 21 cm B) 147 cm² A) 41 cm² C) 273 cm² D) 182 cm² 5)

he area of the shaded region of the figure. Use 3.14 for π as necessary. 4) T 10 in. 30 in. 4= 1/2 bh ↓ A=TT(192) V ½/2 (30) (10)78.5 150 4) 15 in. ㅓ 1.5 A) 186 in.2 B) 146 in.2 C) 221 in.2 D) 71 in.2 Find the area. Use 3.14 for л as necessary. 5) Lookover 7 cm A=1/2bh -1/2(7.21) 13 cm (73.5) A = 1/2 (πr²) 12 (3.14-132) ↓ 265.33 265.33-735 = 191.83 ?!! 21 cm B) 147 cm² A) 41 cm² C) 273 cm² D) 182 cm² 5)

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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he area of the shaded region of the figure. Use 3.14 for π as necessary. 4) T 10 in. 30 in. 4= 1/2 bh ↓ A=TT(192) V ½/2 (30) (10)78.5 150 4) 15 in. ㅓ 1.5 A) 186 in.2 B) 146 in.2 C) 221 in.2 D) 71 in.2 Find the area. Use 3.14 for л as necessary. 5) Lookover 7 cm A=1/2bh -1/2(7.21) 13 cm (73.5) A = 1/2 (πr²) 12 (3.14-132) ↓ 265.33 265.33-735 = 191.83 ?!! 21 cm B) 147 cm² A) 41 cm² C) 273 cm² D) 182 cm² 5)
Transcribed Image Text:he area of the shaded region of the figure. Use 3.14 for π as necessary. 4) T 10 in. 30 in. 4= 1/2 bh ↓ A=TT(192) V ½/2 (30) (10)78.5 150 4) 15 in. ㅓ 1.5 A) 186 in.2 B) 146 in.2 C) 221 in.2 D) 71 in.2 Find the area. Use 3.14 for л as necessary. 5) Lookover 7 cm A=1/2bh -1/2(7.21) 13 cm (73.5) A = 1/2 (πr²) 12 (3.14-132) ↓ 265.33 265.33-735 = 191.83 ?!! 21 cm B) 147 cm² A) 41 cm² C) 273 cm² D) 182 cm² 5)
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