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SAT II Math II : 3-Dimensional Geometry

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Example Questions

Example Question #381 : Sat Subject Test In Math Ii

One cubic meter is equal to one thousand liters.

A circular swimming pool is $\displaystyle d$ meters in diameter and $\displaystyle f$ meters deep throughout. How many liters of water does it hold?

Possible Answers:

$1,000\pi d^{2}f$ $\displaystyle 1,000\pi d^{2}f$

$500\pi df$ $\displaystyle 500\pi df$

$250\pi d^{2}f$ $\displaystyle 250\pi d^{2}f$

$500 \pi d^{2}f$ $\displaystyle 500 \pi d^{2}f$

$1,000\pi df$ $\displaystyle 1,000\pi df$

Correct answer:

$250\pi d^{2}f$ $\displaystyle 250\pi d^{2}f$

Explanation:

The pool can be seen as a cylinder with depth (or height) h= f $\displaystyle h= f$, and a base with diameter $\displaystyle d$ - and, subsequently, radius half this, or $\frac{d}{2}$ $\displaystyle \frac{d}{2}$. The volume of the pool in cubic meters is

$V = \pi r^{2}h = \pi \left ( \frac{d}{2} \right )^{2} \cdot f = \pi \cdot \frac{d^{2}}{4} \cdot f = \frac{ \pi d^{2}f}{4}$ $\displaystyle V = \pi r^{2}h = \pi \left ( \frac{d}{2} \right )^{2} \cdot f = \pi \cdot \frac{d^{2}}{4} \cdot f = \frac{ \pi d^{2}f}{4}$

Multiply this number of cubic meters by 1,000 liters per cubic meter:

$V = \frac{ \pi d^{2}f}{4} \cdot \1,000 = 250\pi d^{2}f$ $\displaystyle V = \frac{ \pi d^{2}f}{4} \cdot \1,000 = 250\pi d^{2}f$

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Example Question #1 : 3 Dimensional Geometry

A water tank takes the shape of a sphere whose exterior has radius 16 feet; the tank is three inches thick throughout. To the nearest hundred, give the surface area of the interior of the tank in square feet.

Possible Answers:

$3,100 \textrm{ ft}^{2}$ $\displaystyle 3,100 \textrm{ ft}^{2}$

$15,600 \textrm{ ft}^{2}$ $\displaystyle 15,600 \textrm{ ft}^{2}$

$16,400 \textrm{ ft}^{2}$ $\displaystyle 16,400 \textrm{ ft}^{2}$

$3,000 \textrm{ ft}^{2}$ $\displaystyle 3,000 \textrm{ ft}^{2}$

$17,100 \textrm{ ft}^{2}$ $\displaystyle 17,100 \textrm{ ft}^{2}$

Correct answer:

$3,100 \textrm{ ft}^{2}$ $\displaystyle 3,100 \textrm{ ft}^{2}$

Explanation:

Three inches is equal to 0.25 feet, so the radius of the interior of the tank is

$\displaystyle 16 - 0.25 = 15.75$ feet.

The surface area of the interior of the tank can be calculated using the formula

$SA = 4 \pi r^{2}$ $\displaystyle SA = 4 \pi r^{2}$

$SA \approx 4 \cdot 3.14 \cdot 15.75 ^{2} \approx 3,116$ $\displaystyle SA \approx 4 \cdot 3.14 \cdot 15.75 ^{2} \approx 3,116$,

which rounds to 3,100 square feet.

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Example Question #383 : Sat Subject Test In Math Ii

A water tank takes the shape of a closed rectangular prism whose exterior has height 30 feet, length 20 feet, and width 15 feet. Its walls are one foot thick throughout. How many cubic feet of water does the tank hold?

Possible Answers:

$7,714\textup{ ft}^{3}$ $\displaystyle 7,714\textup{ ft}^{3}$

$2,446\textup{ ft}^{3}$ $\displaystyle 2,446\textup{ ft}^{3}$

$6,552\textup{ ft}^{3}$ $\displaystyle 6,552\textup{ ft}^{3}$

$10,416\textup{ ft}^{3}$ $\displaystyle 10,416\textup{ ft}^{3}$

$2,204\textup{ ft}^{3}$ $\displaystyle 2,204\textup{ ft}^{3}$

Correct answer:

$6,552\textup{ ft}^{3}$ $\displaystyle 6,552\textup{ ft}^{3}$

Explanation:

The height, length, and width of the interior tank are each two feet less than the corresponding dimension of the exterior of the tank, so the dimensions of the interior are 28, 18, and 13 feet. Multiply these to get the volume:

$V = 28 \cdot 18 \cdot 13 = 6,552$ $\displaystyle V = 28 \cdot 18 \cdot 13 = 6,552$ cubic feet.

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Example Question #384 : Sat Subject Test In Math Ii

A circular swimming pool has diameter 40 meters and depth $\displaystyle t$ meters throughout. Which of the following expressions gives the amount of water it holds, in cubic meters?

Possible Answers:

$80 \pi t$ $\displaystyle 80 \pi t$

$1,600 \pi t$ $\displaystyle 1,600 \pi t$

$400 \pi t$ $\displaystyle 400 \pi t$

$160 \pi t$ $\displaystyle 160 \pi t$

$40 \pi t$ $\displaystyle 40 \pi t$

Correct answer:

$400 \pi t$ $\displaystyle 400 \pi t$

Explanation:

The pool can be seen as a cylinder with depth (or height) h= t $\displaystyle h= t$, and a base with diameter 40 m - and radius half this, or r= 20 $\displaystyle r= 20$. The capacity of the pool is the volume of this cylinder, which is

$V = \pi r^{2}h = \pi \cdot 20^{2}\cdot t = 400 \pi t$ $\displaystyle V = \pi r^{2}h = \pi \cdot 20^{2}\cdot t = 400 \pi t$

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Example Question #385 : Sat Subject Test In Math Ii

One cubic meter is equal to one thousand liters.

A rectangular swimming pool is $\displaystyle D$ meters deep throughout and $\displaystyle W$ meters wide. Its length is ten meters greater than twice its width. How many liters of water does the pool hold?

Possible Answers:

$0.002 DW^{2} + 0.01 DW$ $\displaystyle 0.002 DW^{2} + 0.01 DW$

$2,000 DW^{2} -10,000 DW$ $\displaystyle 2,000 DW^{2} -10,000 DW$

$0.002 DW^{2} - 0.01 DW$ $\displaystyle 0.002 DW^{2} - 0.01 DW$

$2,000 DW^{2} + 10,000 DW$ $\displaystyle 2,000 DW^{2} + 10,000 DW$

None of the other responses is correct.

Correct answer:

$2,000 DW^{2} + 10,000 DW$ $\displaystyle 2,000 DW^{2} + 10,000 DW$

Explanation:

Since the length of the pool is ten meters longer than twice its width $\displaystyle W$, its length is 2W + 10 $\displaystyle 2W + 10$.

The inside of the pool can be seen as a rectangular prism, and as such, its volume in cubic feet can be calculated as the product of its length, width, and height (or depth). This product is

$V = D \cdot (2W + 10) \cdot W$ $\displaystyle V = D \cdot (2W + 10) \cdot W$

$V = 2DW^{2} + 10DW$ $\displaystyle V = 2DW^{2} + 10DW$

Multiply this by the conversion factor 1,000, and its volume in liters is

$V =1,000 \left ( 2DW^{2} + 10DW \right )$ $\displaystyle V =1,000 \left ( 2DW^{2} + 10DW \right )$

$V = 2,000 DW^{2} + 10,000 DW$ $\displaystyle V = 2,000 DW^{2} + 10,000 DW$

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Example Question #386 : Sat Subject Test In Math Ii

A circular swimming pool has diameter 80 feet and depth five feet throughout. To the nearest thousand, how many gallons of water does it hold?

Use the conversion factor: One cubic foot = 7.5 gallons.

Possible Answers:

$188,000\textrm{ gal}$ $\displaystyle 188,000\textrm{ gal}$

$3,000\textrm{ gal}$ $\displaystyle 3,000\textrm{ gal}$

$94,000\textrm{ gal}$ $\displaystyle 94,000\textrm{ gal}$

$754,000\textrm{ gal}$ $\displaystyle 754,000\textrm{ gal}$

$5,000\textrm{ gal}$ $\displaystyle 5,000\textrm{ gal}$

Correct answer:

$188,000\textrm{ gal}$ $\displaystyle 188,000\textrm{ gal}$

Explanation:

The pool can be seen as a cylinder with depth (or height) 5 feet, and a base with diameter 80 feet - and radius half this, or 40 feet. The capacity of the pool is the volume of this cylinder, which is

$V = \pi r ^{2} h \approx 3.14 \times 40^{2} \times 5 \approx 25,120$ $\displaystyle V = \pi r ^{2} h \approx 3.14 \times 40^{2} \times 5 \approx 25,120$ cubic feet.

One cubic foot is equal to 7.5 gallons, so multiply:

$25,120 \times 7.5 = 188,400$ $\displaystyle 25,120 \times 7.5 = 188,400$ gallons

This rounds to 188,000 gallons.

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Example Question #387 : Sat Subject Test In Math Ii

Pool

The above depicts a rectangular swimming pool for an apartment.

On the left and right edges, the pool is three feet deep; the dashed line at the very center represents the line along which it is eight feet deep. Going from the left to the center, its depth increases uniformly; going from the center to the right, its depth decreases uniformly.

In cubic feet, how much water does the pool hold?

Possible Answers:

$9,625 \textrm{ ft}^{3}$ $\displaystyle 9,625 \textrm{ ft}^{3}$

$10,225 \textrm{ ft}^{3}$ $\displaystyle 10,225 \textrm{ ft}^{3}$

$2,070 \textrm{ ft}^{3}$ $\displaystyle 2,070 \textrm{ ft}^{3}$

$9,875 \textrm{ ft}^{3}$ $\displaystyle 9,875 \textrm{ ft}^{3}$

$1,870 \textrm{ ft}^{3}$ $\displaystyle 1,870 \textrm{ ft}^{3}$

Correct answer:

$9,625 \textrm{ ft}^{3}$ $\displaystyle 9,625 \textrm{ ft}^{3}$

Explanation:

The pool can be looked at as a pentagonal prism with "height" 35 feet and its bases the following shape (depth exaggerated):

Pool

This is a composite of two trapezoids, each with bases 3 feet and 8 feet and height 25 feet; the area of each is

$A = \frac{1}{2} (3+8) \cdot 25 = 137.5$ $\displaystyle A = \frac{1}{2} (3+8) \cdot 25 = 137.5$ square feet.

The area of the base is twice this, or

$B = 137.5 \cdot 2 = 275$ $\displaystyle B = 137.5 \cdot 2 = 275$ square feet.

The volume of a prism is its height times the area of its base, or

$V = 275 \cdot 35 = 9,625$ $\displaystyle V = 275 \cdot 35 = 9,625$ cubic feet, the capacity of the pool.

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Example Question #8 : Volume

The bottom surface of a rectangular prism has area 100; the right surface has area 200; the rear surface has area 300. Give the volume of the prism (nearest whole unit), if applicable.

Possible Answers:

3,464 $\displaystyle 3,464$

2,449 $\displaystyle 2,449$

1,000 $\displaystyle 1,000$

6,000 $\displaystyle 6,000$

4,243 $\displaystyle 4,243$

Correct answer:

2,449 $\displaystyle 2,449$

Explanation:

Let the dimensions of the prism be $\displaystyle L$, $\displaystyle W$, and $\displaystyle H$.

Then, $\displaystyle LW = 100$, $\displaystyle LH = 200$, and $\displaystyle WH = 300$.

From the first and last equations, dividing both sides, we get

$\displaystyle WH = 300$

$\displaystyle LW = 100$

$\frac{WH}{LW} = \frac{300}{100}$ $\displaystyle \frac{WH}{LW} = \frac{300}{100}$

$\frac{H}{L} = 3$ $\displaystyle \frac{H}{L} = 3$

Along with the second equation, multiply both sides:

$\displaystyle LH = 200$

$\frac{H}{L} \cdot LH = 3 \cdot 200$ $\displaystyle \frac{H}{L} \cdot LH = 3 \cdot 200$

$H ^{2}= 600$ $\displaystyle H ^{2}= 600$

Taking the square root of both sides and simplifying, we get

$H = \sqrt{600} = \sqrt{100} \cdot \sqrt{6} = 10 \sqrt{6}$ $\displaystyle H = \sqrt{600} = \sqrt{100} \cdot \sqrt{6} = 10 \sqrt{6}$

Now, substituting and solving for the other two dimensions:

$\displaystyle LH = 200$

$L \cdot 10 \sqrt{6} = 200$ $\displaystyle L \cdot 10 \sqrt{6} = 200$

$L = \frac{200}{10 \sqrt{6}} = \frac{20}{\sqrt{6}}$ $\displaystyle L = \frac{200}{10 \sqrt{6}} = \frac{20}{\sqrt{6}}$

$\displaystyle WH = 300$

$W \cdot 10 \sqrt{6} = 300$ $\displaystyle W \cdot 10 \sqrt{6} = 300$

$W = \frac{300}{10 \sqrt{6}} = \frac{30 }{\sqrt{6}}$ $\displaystyle W = \frac{300}{10 \sqrt{6}} = \frac{30 }{\sqrt{6}}$

Now, multiply the three dimensions to obtain the volume:

V= LWH $\displaystyle V= LWH$

$= \frac{20}{\sqrt{6}} \cdot 10 \sqrt{6} \cdot \frac{30 }{\sqrt{6}}$ $\displaystyle = \frac{20}{\sqrt{6}} \cdot 10 \sqrt{6} \cdot \frac{30 }{\sqrt{6}}$

$= \frac{20}{\sqrt{6}} \cdot\frac{ 10 \sqrt{6}}{1} \cdot \frac{30 }{\sqrt{6}}$ $\displaystyle = \frac{20}{\sqrt{6}} \cdot\frac{ 10 \sqrt{6}}{1} \cdot \frac{30 }{\sqrt{6}}$

$= \frac{ 6,000 \sqrt{6}} {6}$ $\displaystyle = \frac{ 6,000 \sqrt{6}} {6}$

$= 1,000 \sqrt{6}$ $\displaystyle = 1,000 \sqrt{6}$

$\approx 1,000 \cdot 2.449$ $\displaystyle \approx 1,000 \cdot 2.449$

$\approx 2,449$ $\displaystyle \approx 2,449$

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Example Question #1 : Volume

The width of a box is two-thirds its height and three-fifths its length. The volume of the box is 6 cubic meters. To the nearest centimeter, give the width of the box.

Possible Answers:

$223 \textrm{ cm}$ $\displaystyle 223 \textrm{ cm}$

$201 \textrm{ cm}$ $\displaystyle 201 \textrm{ cm}$

$246 \textrm{ cm}$ $\displaystyle 246 \textrm{ cm}$

$134 \textrm{ cm}$ $\displaystyle 134 \textrm{ cm}$

$164 \textrm{ cm}$ $\displaystyle 164 \textrm{ cm}$

Correct answer:

$134 \textrm{ cm}$ $\displaystyle 134 \textrm{ cm}$

Explanation:

Call $\displaystyle L$, $\displaystyle H$, and $\displaystyle W$ the length, width, and height of the crate.

The width is two-thirds the height, so

$W = \frac{2}{3} H$ $\displaystyle W = \frac{2}{3} H$.

Equivalently,

$\frac{3} {2} \cdot \frac{2}{3} H = \frac{3} {2} W$ $\displaystyle \frac{3} {2} \cdot \frac{2}{3} H = \frac{3} {2} W$

$H = \frac{3} {2} W$ $\displaystyle H = \frac{3} {2} W$

The width is three-fifths the length, so

$W = \frac{3}{5} L$ $\displaystyle W = \frac{3}{5} L$.

Equivalently,

$\frac{5} {3} \cdot \frac{3}{5} L = \frac{5} {3} \cdot W$ $\displaystyle \frac{5} {3} \cdot \frac{3}{5} L = \frac{5} {3} \cdot W$

$L = \frac{5} {3 } W$ $\displaystyle L = \frac{5} {3 } W$

The dimensions of the crate in terms of $\displaystyle W$ are $\displaystyle W$, $\frac{3} {2} W$ $\displaystyle \frac{3} {2} W$, and $\frac{5} {3 } W$ $\displaystyle \frac{5} {3 } W$. The volume is their product:

V= LWH $\displaystyle V= LWH$,

Substitute:

LWH = V $\displaystyle LWH = V$

$\frac{5} {3 } W \cdot W \cdot \frac{3} {2} W = V$ $\displaystyle \frac{5} {3 } W \cdot W \cdot \frac{3} {2} W = V$

$\frac{5} {3 } \cdot \frac{3} {2} \cdot W \cdot W \cdot W = 6$ $\displaystyle \frac{5} {3 } \cdot \frac{3} {2} \cdot W \cdot W \cdot W = 6$

$\frac{5} {2} \cdot W ^{3}= 6$ $\displaystyle \frac{5} {2} \cdot W ^{3}= 6$

$\frac{2} {5} \cdot \frac{5} {2} \cdot W ^{3}= \frac{2} {5} \cdot 6$ $\displaystyle \frac{2} {5} \cdot \frac{5} {2} \cdot W ^{3}= \frac{2} {5} \cdot 6$

$W ^{3}= \frac{12} {5}$ $\displaystyle W ^{3}= \frac{12} {5}$

Taking the cube root of both sides:

$W =\sqrt[3]{ \frac{12} {5}}$ $\displaystyle W =\sqrt[3]{ \frac{12} {5}}$

$W \approx 1.339$ $\displaystyle W \approx 1.339$ meters.

Since one meter comprises 100 centimeters, multiply by 100 to convert to centimeters:

$1.339 \cdot 100 = 133.9$ $\displaystyle 1.339 \cdot 100 = 133.9$ centimeters,

which rounds to 134 centimeters.

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Example Question #10 : Volume

Box 2

The shaded face of the rectangular prism in the above diagram is a square. The volume of the prism is $\displaystyle V$; give the value of $\displaystyle x$ in terms of $\displaystyle V$.

Possible Answers:

$x =\frac{ \sqrt{V}}{25}$ $\displaystyle x =\frac{ \sqrt{V}}{25}$

$x =\frac{ V}{625}$ $\displaystyle x =\frac{ V}{625}$

$x =\frac{ \sqrt{V}}{5}$ $\displaystyle x =\frac{ \sqrt{V}}{5}$

$x =\frac{ V}{25}$ $\displaystyle x =\frac{ V}{25}$

$x =\frac{ \sqrt{V}}{625}$ $\displaystyle x =\frac{ \sqrt{V}}{625}$

Correct answer:

$x =\frac{ \sqrt{V}}{5}$ $\displaystyle x =\frac{ \sqrt{V}}{5}$

Explanation:

The volume of a rectangular prism is the product of its length, its width, and its height; that is,

LWH = V $\displaystyle LWH = V$

Since the shaded face of the prism is a square, we can set $\displaystyle L = H = x$, and W = 25 $\displaystyle W = 25$; substituting and solving for $\displaystyle x$:

$x \cdot 25 \cdot x = V$ $\displaystyle x \cdot 25 \cdot x = V$

$25x^{2} = V$ $\displaystyle 25x^{2} = V$

$\frac{25x^{2}}{25} = \frac{V}{25}$ $\displaystyle \frac{25x^{2}}{25} = \frac{V}{25}$

$x^{2} = \frac{V}{25}$ $\displaystyle x^{2} = \frac{V}{25}$

Taking the positive square root of both sides, and simplifying the expression on the right using the Quotient of Radicals Rule:

$x =\sqrt{ \frac{V}{25}} = \frac{\sqrt{V}}{\sqrt{25}} = \frac{\sqrt{V}}{5}$ $\displaystyle x =\sqrt{ \frac{V}{25}} = \frac{\sqrt{V}}{\sqrt{25}} = \frac{\sqrt{V}}{5}$

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SAT II Math II : 3-Dimensional Geometry

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #381 : Sat Subject Test In Math Ii

Example Question #1 : 3 Dimensional Geometry

Example Question #383 : Sat Subject Test In Math Ii

Example Question #384 : Sat Subject Test In Math Ii

Example Question #385 : Sat Subject Test In Math Ii

Example Question #386 : Sat Subject Test In Math Ii

Example Question #387 : Sat Subject Test In Math Ii

Example Question #8 : Volume

Example Question #1 : Volume

Example Question #10 : Volume

All SAT II Math II Resources

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