# Chapter 10 Test Review

```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 = 4/3r3
But since it is a hemisphere, we
have to divide by 2.
Therefore, the Volume = 2/3r3
r= 6.5, So V = 2/3(6.53) = 575.2 in3
Discovering Geometry Chapter 10 Test Review
HGSH
Volume = 4/3r3
4500 = 4/3r3
4500 = 4/3r3
4500(3/4) = r3
15 = r
Surface Area = 4r2 = 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= 1/3r2(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
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