Discovering Geometry Chapter 10 Test Review HGSH Volume of one cube is: (2x)(2x)(x) = 4x3 and there are 5 cubes. The volume = 5(4x3) = 20x3 x = 7, The volume of the cross = 20x3 =20(73) = 6860 cm3 Discovering Geometry Chapter 10 Test Review HGSH If the rectangular face area = 375m2, then 375/15 = 25m. So the hypotenuse of the triangle = 25m. Using the Pythagorean Theorem, then the second leg of the triangle is: c2 – a2 = b2 , 252 – 202 = b2 , 625 – 400 = b2 , 225 = b2, So b=15m. Now the area of the triangle is ½(b)(h), where b=15 and h=20, B= ½(15)(20) = 150m2 The Volume of the Prism = BH where B= 150, and H=15, V=BH = (150)(15)= 2250 m3 Discovering Geometry Chapter 10 Test Review HGSH The area of the triangle is ½(b)(h), where b=1 and h=2, B= ½(1)(2) = 1m2 The Volume of the Prism = BH where B= 1m2, and H=13m, V=BH = (1)(13)= 13m3 Discovering Geometry Chapter 10 Test Review HGSH The area of the triangle is ½(b)(h), where b=1 and h=2, B= ½(1)(2) = 1m2 The Volume of the Prism = BH where B= 1m2, and H=15m, V=BH = (1)(15)= 15m3 Discovering Geometry Chapter 10 Test Review HGSH The Volume for the Hexagonal Pyramid is: 1/3BH, where B= 76.8 in2 and H=18.1 in The Volume = 1/3BH = 1/3(76.8)(18.1) in3 = 463.4 in3 Discovering Geometry Chapter 10 Test Review HGSH Compare the volumes: V=BH Vol. of Box = (12)(5)(18) = 1080 cm3 Vol. Of Cyl. = (r2)(18) = (52)(18) = 1413.7 cm3 The Cylinder holds the greater amount of cereal. 1413.7 cm3 Discovering Geometry Chapter 10 Test Review HGSH Amount of Water Displaced 3 in. 24 in. 14 in. Volume of Water Displaced: (14)(24)(3) = 1008 in3 Since 85 rocks were added to the tank, then: 1008/85 = 11.9 in3 Discovering Geometry Chapter 10 Test Review HGSH Volume = 4/3r3 But since it is a hemisphere, we have to divide by 2. Therefore, the Volume = 2/3r3 r= 6.5, So V = 2/3(6.53) = 575.2 in3 Discovering Geometry Chapter 10 Test Review HGSH Volume = 4/3r3 4500 = 4/3r3 4500 = 4/3r3 4500(3/4) = r3 15 = r Surface Area = 4r2 = 4(152) = 900 in2 Discovering Geometry Chapter 10 Test Review HGSH Volume = (11)(6)(2) = 132m3 Discovering Geometry Chapter 10 Test Review HGSH Volume = BH = (r2)(H) = (22)(11) = 44 in3 Discovering Geometry Chapter 10 Test Review HGSH Volume = BH = (r2)(H) = (32)(24) = 216 cm3 Discovering Geometry Chapter 10 Test Review HGSH Volume = 1/3BH = 1/3(12)(12)(11) = 528m3 Discovering Geometry Chapter 10 Test Review HGSH 20 in. 22 in. 15 in. Volume = BH = (15)(22)(20) = 6600in3 1 Gallon [Fluid, US] = 231 Cubic Inches 6600/231 = 28.6 gallons. Discovering Geometry Chapter 10 Test Review HGSH Volume = 1/3BH , B=r2 , H=? 800= 1/3r2(H) 20 Ft. 800(3)/(10)2 = (H) 800(3)/100 = (H) 8(3)/ = H 24/ = H 24/ = H 7.6 Ft = H Discovering Geometry Chapter 10 Test Review HGSH Amount of Water Displaced 0.5 Ft. Volume of Water Displaced: V = (6)(4)(0.5) = 12 Ft3 4 Ft. 6 Ft. Discovering Geometry Chapter 10 Test Review HGSH Density = Mass / Volume Volume = Mass / Density V = (36.5)/(0.78) = 46.8 cm3 Discovering Geometry Chapter 10 Test Review HGSH Discovering Geometry Chapter 10 Test Review HGSH

© Copyright 2018