Discovering Geometry Chapter 10 Test Review...

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Discovering Geometry Chapter 10 Test Review HGSH

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) = 20x

3

x = 7, The volume of the cross = 20x

3 =20(7

3) = 6860 cm

3

Discovering Geometry Chapter 10 Test Review HGSH

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

Using the Pythagorean Theorem, then the

second leg of the triangle is:

c2 – a

2 = b

2 , 25

2 – 20

2 = b

2 , 625 – 400 = b

2 ,

225 = b2, So b=15m.

If the rectangular face area = 375m2, then

375/15 = 25m. So the hypotenuse of the

triangle = 25m.

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)= 13m

3

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)= 15m

3

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) = (5

2)(18)

= 1413.7 cm3

The Cylinder holds the greater

amount of cereal. 1413.7 cm3

Discovering Geometry Chapter 10 Test Review HGSH

3 in.

24 in.

14 in.

Amount of Water Displaced

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 in

3

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(15

2) = 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) = (2

2)(11) = 44 in3

Discovering Geometry Chapter 10 Test Review HGSH

Volume = BH = (r2)(H) = (3

2)(24) = 216 cm

3

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

20 Ft.

Volume = 1/3BH , B=r2 , H=?

800= 1/3r2(H)

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

0.5 Ft.

4 Ft.

6 Ft.

Amount of Water Displaced

Volume of Water Displaced:

V = (6)(4)(0.5) = 12 Ft3

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