Discovering Geometry Chapter 10 Test Review...

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Discovering Geometry Chapter 10 Test Review HGSH

Transcript of Discovering Geometry Chapter 10 Test Review...

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

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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

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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