Surface Area and Volume Answers to CH. 11 Exam

Find the volume

• To find the volume, first figure out the shapes involved

𝑉 𝑐𝑜𝑛𝑒+𝑉 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟+𝑉 h h𝑒𝑚𝑖𝑠𝑝 𝑒𝑟𝑒

Find the volume

𝑉 𝑐𝑜𝑛𝑒+𝑉 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟+𝑉 h h𝑒𝑚𝑖𝑠𝑝 𝑒𝑟𝑒

13𝜋 𝑟2h+𝜋𝑟 2h+

23𝜋𝑟 3

13𝜋 52∙5+𝜋 52 ∙5+

23𝜋 53

1253

𝜋+125𝜋+2503

𝜋

250𝜋≈785.4

Find the surface area

Figure out the parts of the shapes involved then add them together.

𝐿 . 𝐴 .𝑐𝑜𝑛𝑒+𝐿 . 𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟+𝐿 . 𝐴 .h h𝑒𝑚𝑖𝑠𝑝 𝑒𝑟𝑒

Find the surface area

𝐿 . 𝐴 .𝑐𝑜𝑛𝑒+𝐿 . 𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟+𝐿 . 𝐴 .h h𝑒𝑚𝑖𝑠𝑝 𝑒𝑟𝑒

𝜋𝑟𝑙+2𝜋 h𝑟 +2𝜋𝑟 2❑𝜋 ∙5 ∙√50+2 ∙𝜋 ∙5 ∙5+2 ∙𝜋 ∙52

25√2𝜋+50𝜋+50𝜋25√2𝜋+100𝜋≈ 425.23

To find surface area, you do not include the circles, just the outside of the composite shape.

Find the volume

To find the volume, form break the composite figure into three prisms and add together

Find the volume

𝑉 𝑙𝑎𝑟𝑔𝑒+𝑉𝑚𝑒𝑑𝑖𝑢𝑚+𝑉 𝑠𝑚𝑎𝑙𝑙

12

44

9

3

4

43

5

(12 ∙4 ∙ 4)+(9 ∙3 ∙ 4)+(5 ∙3 ∙ 4)192+108+60𝑽=𝟑𝟔𝟎

Find the surface area

To find the surface area of the figure you have to break the composite figure into rectangles and add their areas together

Find the surface area

back

4

12

bottom

4

10

front

45

4

3

side

4 3 3

129

5

top

44

3

3

¿ (12 ∙4 )+¿(12 ∙4 )+ (9 ∙3 )+(5 ∙3 )+¿(5 ∙4 )+(4 ∙4 )+(4 ∙3 )+¿(10 ∙4 )+¿(4 ∙4 )+(3 ∙3 )+(3 ∙3)¿𝟒𝟖+𝟐 (𝟗𝟎)+𝟒𝟖+𝟒𝟎+𝟒𝟎¿𝟑𝟓𝟔

Find the volume

To find the volume of the figure you have to break apart the composite figure

Find the volume

Half cylinder with radius 3.5

Rectangular prism

¿12𝑉 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 1+

12𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 2+𝑉 𝑝𝑟𝑖𝑠𝑚

¿12

(𝜋 h𝑟 )+ 12

(𝜋 h𝑟 )+( h𝑙𝑤 )

¿12

(𝜋 3.5∙8 )+ 12

(𝜋 3.5 ∙4 )+(8 ∙4 ∙7)

≈𝟒𝟓𝟒 .𝟗

Find the surface area

To find the surface area of the figure you have to break apart the composite figure

Find the surface area

Half cylinder with radius 3.5

Rectangular prism without the top or the side

¿12𝑆 .𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 1+

12𝑆 .𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 2+𝑆 . 𝐴 .𝑝𝑟𝑖𝑠𝑚

¿12

(2𝜋 ∙3.5 ∙8 )+2𝜋 3.52+ 12

(2𝜋 ∙3.5 ∙4 )+2𝜋 3.52+2 (8 ∙ 4 )+(4 ∙7 )+(8 ∙7)

¿12

(2𝜋 h𝑟 +2𝜋 𝑟2 )+ 12

(2𝜋 h𝑟 +2𝜋 𝑟2 )+2 ( 𝑙𝑤 𝑓𝑟𝑜𝑛𝑡 )+(𝑙𝑤𝑠𝑖𝑑𝑒 )+ (𝑙𝑤𝑏𝑜𝑡𝑡𝑜𝑚 )

≈𝟑𝟓𝟔 .𝟗𝟐

Use algebra to express volume

To solve algebraically, first identify the shapes and the dimensions given.

Use algebra to express volumeWhat we know.

𝑟𝑎𝑑𝑖𝑢𝑠=2𝑎h h𝑒𝑖𝑔 𝑡=𝑏¿13𝜋 ¿

¿ 4 𝑎2𝑏𝜋3

¿ 𝟖𝒂𝟐𝒃𝝅𝟑

The volume of the solid is 360 cubic feet. Find the value of .

First use the Pythagorean Theorem to find the missing side of the triangle

𝑎2+𝑏2=𝑐2

𝑎2+64=289𝑎𝑚𝑖𝑠𝑠𝑖𝑛𝑔 𝑠𝑖𝑑𝑒=15

Now solve for 𝑉 𝑝𝑟𝑖𝑠𝑚= h𝐵360=

12∙15 ∙8∙ 𝑥

360=60 𝑥𝟔=𝒙

Find the volume

Find the volume of the solid of revolution rotated about the line.

Find the volume of the solid of revolution rotated about the line.

𝑉=𝑉 𝑙𝑎𝑟𝑔𝑒−𝑉 𝑠𝑚𝑎𝑙𝑙

𝑉=𝜋 𝑟2h−𝜋𝑟 2h𝑉=𝜋 ∙42 ∙4−𝜋 ∙22 ∙4𝑉=64𝜋−16𝜋𝑽=𝟒𝟖 𝝅𝑽 ≈𝟏𝟓𝟎 .𝟖

Find the total surface area of the solid of revolution found by rotating the triangle about the vertical line

What does the solid look like?

15

8

Now that we know what the solid looks like, we realize that the surface area is what is covered by the green.

What is the surface area covered by green?

𝑆 . 𝐴 .=𝐴𝑟𝑒𝑎𝑐𝑖𝑟𝑐𝑙𝑒+𝐿 .𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟+𝐿 . 𝐴 .𝑐𝑜𝑛𝑒¿𝜋𝑟 2+2𝜋 h𝑟 +𝜋𝑟𝑙¿𝜋 ∙152+2𝜋 ∙15 ∙8+𝜋 ∙15 ∙17¿225𝜋+240𝜋+255𝜋¿𝟕𝟐𝟎𝝅¿𝟐𝟐𝟔𝟏 .𝟓𝝅

𝑟=15h=8𝑙=15

Solid I is similar to Solid II, find the value of .

First find the ratio of the solids.

or

𝟏𝟑

=𝑥−22𝑥+3

Set up your proportion and solve

2 𝑥+3=3𝑥−6𝒙=𝟗

What is the height of a cone whose slant height is twice the radius and whose volume is

𝑟

h2𝑟h2+𝑟2=(2𝑟 )2

h2+𝑟2=4𝑟 2

𝒉=√𝟑𝒓❑

first find h

Plug in and solve

𝑉=13𝜋𝑟2h=343𝜋 √3

2413𝜋 𝑟2𝑟 √3=343𝜋√3

24

𝜋𝑟3√33

=343𝜋 √324𝒓=

𝟕𝟐𝜋𝑟3

3=343𝜋 √3

24 √3𝑟3

3=343𝜋24𝜋

𝑟3=3 ∙34324

𝑟3=3438

1

8

h=𝑟 √3𝒉=

𝟕𝟐

√𝟑

Bathman just discovered that the valve on his cement truck (why he has a cement truck I don’t know), failed during the night and that all the contents ran out to form a giant cone of hardened cement. To make an insurance claim, Bathman needs to figure out how much cement is in the cone. The circumference of its base is 44 feet and it is 5 feet high. Calculate the volume.

Use circumference to find

C=2 π 𝑟4 4=2π 𝑟𝟐𝟐𝝅

=𝒓 ≈𝟕

Now that you have r, plug in an solve

𝑉 𝑐𝑜𝑛𝑒=13π 𝑟2h

𝑉 𝑐𝑜𝑛𝑒=13π ∙72 ∙5

𝑽 𝒄𝒐𝒏𝒆=𝟐𝟒𝟓𝟑

𝛑≈𝟐𝟓𝟕

The solid below is a peg with a square hole. Find its surface area.

In order to find the surface area we will have to:

Find the lateral area of the cylinder

Find the area of the circle minus the square for the top and bottom

Find the lateral area of the square peg

In order to find the surface area we will have to:

Find the lateral area of the cylinder

𝐿 . 𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟=2𝜋 h𝑟𝐿 . 𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟=2 ∙𝜋 ∙7 ∙13𝐿 . 𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟=2 ∙𝜋 ∙7 ∙13𝑳 . 𝑨 .𝒄𝒚𝒍𝒊𝒏𝒅𝒆𝒓=𝟏𝟖𝟐𝝅

Next, we will have to:

Find the area of the circle minus the square for the top and bottom

𝐴𝑡𝑜𝑝 /𝑏𝑜𝑡𝑡𝑜𝑚=𝐴𝑐𝑖𝑟𝑐𝑙𝑒−𝐴𝑠𝑞𝑢𝑎𝑟𝑒¿2(𝜋𝑟2−𝑠2)¿2(𝜋 ∙72−42)≈2(138)≈𝟐𝟕𝟔

Next we will have to:

Find the lateral area of the square peg

𝐿 . 𝐴 .𝑝𝑒𝑔= h𝑃𝐿 . 𝐴 .𝑝𝑒𝑔= (4 ∙ 4 )13

𝑳 . 𝑨 .𝒑𝒆𝒈=𝟐𝟎𝟖

Now we put them all together

+

𝐿 . 𝐴 .𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟=2𝜋 h𝑟

𝐴𝑡𝑜𝑝 /𝑏𝑜𝑡𝑡𝑜𝑚=𝐴𝑐𝑖𝑟𝑐𝑙𝑒−𝐴𝑠𝑞𝑢𝑎𝑟𝑒

𝐿 . 𝐴 .𝑝𝑒𝑔= h𝑃+

𝟏𝟖𝟐𝝅+¿𝟐𝟕𝟔+¿𝟐𝟎𝟖𝑺 . 𝑨 . ≈𝟏𝟎𝟓𝟓 .𝟖

Find the lateral area

• A water bottle in the shape of a cylinder has a volume of 500 cubic centimeters. The diameter of a base is 7.5 cm. What is the approximate height of the bottle?

• Andrew is working on a car and has a funnel with the dimensions shown. He uses the funnel to put oil in his car. Oil flows out of the funnel at a rate of 45 milliliters per second. How long will it take to empty the funnel when it is full of oil?

Find the volume

Find the volume