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High School Math : Solid Geometry

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Example Questions

Example Question #1 : How To Find The Surface Area Of A Polyhedron

Find the surface area of the following polyhedron.

Ice_cream_cone

Possible Answers:

$118 \pi m^2$ $\displaystyle 118 \pi m^2$

$98 \pi m^2$ $\displaystyle 98 \pi m^2$

$88 \pi m^2$ $\displaystyle 88 \pi m^2$

$108 \pi m^2$ $\displaystyle 108 \pi m^2$

$128 \pi m^2$ $\displaystyle 128 \pi m^2$

Correct answer:

$108 \pi m^2$ $\displaystyle 108 \pi m^2$

Explanation:

The formula for the surface area of the polyhedron is:

$SA = SA_{cone}+\frac{1}{2}SA_{sphere}$ $\displaystyle SA = SA_{cone}+\frac{1}{2}SA_{sphere}$

$SA = \pi r l+\frac{1}{2}(4 \pi r^2)$ $\displaystyle SA = \pi r l+\frac{1}{2}(4 \pi r^2)$

$SA = \pi r l+2 \pi r^2$ $\displaystyle SA = \pi r l+2 \pi r^2$

Where $\displaystyle r$ is the radius of the cone, $\displaystyle l$ is the slant height of the cone, and $\displaystyle r$ is the radius of the sphere

Use the formula for a $\displaystyle 30-60-90$ triangle to find the radius and slant height:

$a-a\sqrt{2}-2a$ $\displaystyle a-a\sqrt{2}-2a$

$3\sqrt{3}m-9m-6\sqrt{3}m$ $\displaystyle 3\sqrt{3}m-9m-6\sqrt{3}m$

Plugging in our values, we get:

$SA = \pi r l+2 \pi r^2$ $\displaystyle SA = \pi r l+2 \pi r^2$

$SA = (\pi)(3\sqrt{3}m)(6\sqrt{3}m)+2(\pi)(3\sqrt{3}m)^2$ $\displaystyle SA = (\pi)(3\sqrt{3}m)(6\sqrt{3}m)+2(\pi)(3\sqrt{3}m)^2$

$SA = 54 \pi m^2 + 54 \pi m^2 = 108 \pi m^2$ $\displaystyle SA = 54 \pi m^2 + 54 \pi m^2 = 108 \pi m^2$

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Example Question #3 : Other Polyhedrons

Find the surface area of the following polyhedron.

Dome

Possible Answers:

$81 \pi m^2$ $\displaystyle 81 \pi m^2$

$101 \pi m^2$ $\displaystyle 101 \pi m^2$

$61 \pi m^2$ $\displaystyle 61 \pi m^2$

$91 \pi m^2$ $\displaystyle 91 \pi m^2$

$71 \pi m^2$ $\displaystyle 71 \pi m^2$

Correct answer:

$81 \pi m^2$ $\displaystyle 81 \pi m^2$

Explanation:

The formula for the surface area of the polyhedron is:

$SA = (cone:no\ base)+(cylinder:one\ base)$ $\displaystyle SA = (cone:no\ base)+(cylinder:one\ base)$

$SA = \frac{1}{2}(circumference)(slant\ height)+(base)+(circumference)(height)$ $\displaystyle SA = \frac{1}{2}(circumference)(slant\ height)+(base)+(circumference)(height)$

$SA = \frac{1}{2}(2 \pi r)(h_s)+(\pi r^2)+(2 \pi r)(h)$ $\displaystyle SA = \frac{1}{2}(2 \pi r)(h_s)+(\pi r^2)+(2 \pi r)(h)$

where $\displaystyle r$ is the radius of the cone, h_s $\displaystyle h_s$ is the slant height of the cone, $\displaystyle r$ is the radius of the cylinder, and $\displaystyle h$ is the height of the cylinder.

Use the formula for a $\displaystyle 30-60-90$ triangle to find the length of the radius:

$a-a\sqrt{3}-2a$ $\displaystyle a-a\sqrt{3}-2a$

$3m-3\sqrt{3}m-6m$ $\displaystyle 3m-3\sqrt{3}m-6m$

Plugging in our values, we get:

$SA = \frac{1}{2}(2\pi (3m))(6m)+(\pi)(3m)^2+(2\pi (3m)(9m))$ $\displaystyle SA = \frac{1}{2}(2\pi (3m))(6m)+(\pi)(3m)^2+(2\pi (3m)(9m))$

$SA = 81 \pi m^2$ $\displaystyle SA = 81 \pi m^2$

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Example Question #2 : How To Find The Surface Area Of A Polyhedron

Find the surface area of the following polyhedron.

Ice_cream_cone

Possible Answers:

$64\sqrt{2}\pi + 32 \pi m^2$ $\displaystyle 64\sqrt{2}\pi + 32 \pi m^2$

$32\sqrt{2}\pi + 32 \pi m^2$ $\displaystyle 32\sqrt{2}\pi + 32 \pi m^2$

$64\sqrt{2}\pi + 64 \pi m^2$ $\displaystyle 64\sqrt{2}\pi + 64 \pi m^2$

$32\sqrt{2}\pi + 64 \pi m^2$ $\displaystyle 32\sqrt{2}\pi + 64 \pi m^2$

Correct answer:

$32\sqrt{2}\pi + 64 \pi m^2$ $\displaystyle 32\sqrt{2}\pi + 64 \pi m^2$

Explanation:

The formula for the surface area of a polyhedron is:

$SA = (cone)+ \frac{1}{2}(sphere)$ $\displaystyle SA = (cone)+ \frac{1}{2}(sphere)$

$SA = \frac{1}{2}(2\pi r)(h_s)+ \frac{1}{2}(4\pi r^2)$ $\displaystyle SA = \frac{1}{2}(2\pi r)(h_s)+ \frac{1}{2}(4\pi r^2)$

$SA = \pi rh_s+ 2\pi r^2$ $\displaystyle SA = \pi rh_s+ 2\pi r^2$

where $\displaystyle r$ is the radius of the polyhedron and h_s $\displaystyle h_s$ is the slant height of the cone.

Use the formula for a $\displaystyle 45-45-90$ triangle to find the length of the radius:

$a-a-a\sqrt{2}$ $\displaystyle a-a-a\sqrt{2}$

$4\sqrt{2}m-4\sqrt{2}m-8m$ $\displaystyle 4\sqrt{2}m-4\sqrt{2}m-8m$

Plugging in our values, we get:

$SA = \pi (4\sqrt{2}m)(8m)+ 2\pi (4\sqrt{2}m)^2$ $\displaystyle SA = \pi (4\sqrt{2}m)(8m)+ 2\pi (4\sqrt{2}m)^2$

$SA = 32\sqrt{2}\pi + 64 \pi m^2$ $\displaystyle SA = 32\sqrt{2}\pi + 64 \pi m^2$

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Example Question #21 : Solid Geometry

Find the volume of the following half cylinder.

Half_cylinder

Possible Answers:

$207.5 \pi m^3$ $\displaystyle 207.5 \pi m^3$

$187.5 \pi m^3$ $\displaystyle 187.5 \pi m^3$

$157.5 \pi m^3$ $\displaystyle 157.5 \pi m^3$

$200 \pi m^3$ $\displaystyle 200 \pi m^3$

$150.5 \pi m^3$ $\displaystyle 150.5 \pi m^3$

Correct answer:

$187.5 \pi m^3$ $\displaystyle 187.5 \pi m^3$

Explanation:

The formula for the volume of a half-cylinder is:

$V = \frac{1}{2} (\pi r^2 h)$ $\displaystyle V = \frac{1}{2} (\pi r^2 h)$

where $\displaystyle r$ is the radius of the base and $\displaystyle h$ is the length of the height.

Plugging in our values, we get:

$V = \frac{1}{2} \pi (5m)^2 (15m)$ $\displaystyle V = \frac{1}{2} \pi (5m)^2 (15m)$

$V = 187.5 \pi m^3$ $\displaystyle V = 187.5 \pi m^3$

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Example Question #6 : Other Polyhedrons

Find the volume of the following polyhedron.

Ice_cream_cone

Possible Answers:

$81 \pi m^3+54\sqrt{3} \pi m^3$ $\displaystyle 81 \pi m^3+54\sqrt{3} \pi m^3$

$54 \pi m^3+81\sqrt{3} \pi m^3$ $\displaystyle 54 \pi m^3+81\sqrt{3} \pi m^3$

$81 \pi m^3+81\sqrt{3} \pi m^3$ $\displaystyle 81 \pi m^3+81\sqrt{3} \pi m^3$

$81 \pi m^3+50\sqrt{3} \pi m^3$ $\displaystyle 81 \pi m^3+50\sqrt{3} \pi m^3$

$54 \pi m^3+54\sqrt{3} \pi m^3$ $\displaystyle 54 \pi m^3+54\sqrt{3} \pi m^3$

Correct answer:

$81 \pi m^3+54\sqrt{3} \pi m^3$ $\displaystyle 81 \pi m^3+54\sqrt{3} \pi m^3$

Explanation:

The formula for the volume of the polyhedron is:

$V = V_{cone}+\frac{1}{2}V_{sphere}$ $\displaystyle V = V_{cone}+\frac{1}{2}V_{sphere}$

$V = \frac{\pi r^2 h}{3} +\frac{1}{2} \frac{4\pi r^3}{3}$ $\displaystyle V = \frac{\pi r^2 h}{3} +\frac{1}{2} \frac{4\pi r^3}{3}$

$V = \frac{\pi r^2 h}{3} +\frac{2\pi r^3}{3}$ $\displaystyle V = \frac{\pi r^2 h}{3} +\frac{2\pi r^3}{3}$

Where $\displaystyle r$ is the radius of the cone, $\displaystyle h$ is the height of the cone, and $\displaystyle r$ is the radius of the sphere.

Use the formula for a $\displaystyle 30-60-90$ triangle to find the length of the radius:

$a-a\sqrt{2}-2a$ $\displaystyle a-a\sqrt{2}-2a$

$3\sqrt{3}m-9m-6\sqrt{3}m$ $\displaystyle 3\sqrt{3}m-9m-6\sqrt{3}m$

Plugging in our values, we get:

$V = \frac{\pi (3\sqrt{3}m)^29m}{3}+\frac{2 \pi (3\sqrt{3}m)^3}{3}$ $\displaystyle V = \frac{\pi (3\sqrt{3}m)^29m}{3}+\frac{2 \pi (3\sqrt{3}m)^3}{3}$

$V = \frac{\pi (27m^2)9m}{3}+\frac{2 \pi (81\sqrt{3}m^3)}{3}$ $\displaystyle V = \frac{\pi (27m^2)9m}{3}+\frac{2 \pi (81\sqrt{3}m^3)}{3}$

$V = 81 \pi m^3+54\sqrt{3} \pi m^3$ $\displaystyle V = 81 \pi m^3+54\sqrt{3} \pi m^3$

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Example Question #22 : Solid Geometry

Find the volume of the following polyhedron.

Dome

Possible Answers:

$18 \pi \sqrt{3}m^3 + 81 \pi m^3$ $\displaystyle 18 \pi \sqrt{3}m^3 + 81 \pi m^3$

$24 \pi \sqrt{3}m^3 + 81 \pi m^3$ $\displaystyle 24 \pi \sqrt{3}m^3 + 81 \pi m^3$

$6 \pi \sqrt{3}m^3 + 81 \pi m^3$ $\displaystyle 6 \pi \sqrt{3}m^3 + 81 \pi m^3$

$9 \pi \sqrt{3}m^3 + 81 \pi m^3$ $\displaystyle 9 \pi \sqrt{3}m^3 + 81 \pi m^3$

$27 \pi \sqrt{3}m^3 + 81 \pi m^3$ $\displaystyle 27 \pi \sqrt{3}m^3 + 81 \pi m^3$

Correct answer:

$9 \pi \sqrt{3}m^3 + 81 \pi m^3$ $\displaystyle 9 \pi \sqrt{3}m^3 + 81 \pi m^3$

Explanation:

The formula for the volume of the polyhedron is:

$\displaystyle V = (cone)+(cylinder)$

$V = \frac{\pi r^2 h}{3}+\pi r^2 h$ $\displaystyle V = \frac{\pi r^2 h}{3}+\pi r^2 h$

where $\displaystyle r$ is the radius of the cone, $\displaystyle h$ is the height of the cone, $\displaystyle r$ is the radius of the cylinder, and $\displaystyle h$ is the height of the cylinder.

Use the formula for a $\displaystyle 30-60-90$ triangle to find the length of the radius and height of the cone:

$a-a\sqrt{3}-2a$ $\displaystyle a-a\sqrt{3}-2a$

$3m-3\sqrt{3}m-6m$ $\displaystyle 3m-3\sqrt{3}m-6m$

Plugging in our values, we get:

$V = \frac{\pi (3m)^2 (3\sqrt{3}m)}{3}+\pi (3m)^2 (9m)$ $\displaystyle V = \frac{\pi (3m)^2 (3\sqrt{3}m)}{3}+\pi (3m)^2 (9m)$

$V = 9 \pi \sqrt{3}m^3 + 81 \pi m^3$ $\displaystyle V = 9 \pi \sqrt{3}m^3 + 81 \pi m^3$

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Example Question #8 : Other Polyhedrons

Find the volume of the following polyhedron.

Ice_cream_cone

Possible Answers:

$128 \sqrt{2} \pi m^3$ $\displaystyle 128 \sqrt{2} \pi m^3$

$108 \sqrt{2} \pi m^3$ $\displaystyle 108 \sqrt{2} \pi m^3$

$118 \sqrt{2} \pi m^3$ $\displaystyle 118 \sqrt{2} \pi m^3$

$138 \sqrt{2} \pi m^3$ $\displaystyle 138 \sqrt{2} \pi m^3$

$148 \sqrt{2} \pi m^3$ $\displaystyle 148 \sqrt{2} \pi m^3$

Correct answer:

$128 \sqrt{2} \pi m^3$ $\displaystyle 128 \sqrt{2} \pi m^3$

Explanation:

The formula for the volume of the polyhedron is:

$V = (cone) + \frac{1}{2}(sphere)$ $\displaystyle V = (cone) + \frac{1}{2}(sphere)$

$V = \frac{1}{3}(\pi r^2)(h)+\frac{1}{2}\left(\frac{4\pi r^3}{3}\right)$ $\displaystyle V = \frac{1}{3}(\pi r^2)(h)+\frac{1}{2}\left(\frac{4\pi r^3}{3}\right)$

where $\displaystyle r$ is the radius of the polyhedron and $\displaystyle h$ is the height of the cone.

Use the formula for a $\displaystyle 45-45-90$ triangle to find the length of the radius and height:

$a-a-a\sqrt{2}$ $\displaystyle a-a-a\sqrt{2}$

$4\sqrt{2}m-4\sqrt{2}m-8m$ $\displaystyle 4\sqrt{2}m-4\sqrt{2}m-8m$

Plugging in our values, we get:

$V = \frac{1}{3}(\pi (4\sqrt{2}m)^2)(4\sqrt{2}m)+\frac{1}{2}\left(\frac{4\pi (4\sqrt{2}m)^3}{3}\right)$ $\displaystyle V = \frac{1}{3}(\pi (4\sqrt{2}m)^2)(4\sqrt{2}m)+\frac{1}{2}\left(\frac{4\pi (4\sqrt{2}m)^3}{3}\right)$

$V = 128 \sqrt{2} \pi m^3$ $\displaystyle V = 128 \sqrt{2} \pi m^3$

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Example Question #643 : Geometry

Our backyard pool holds 10,000 gallons. Its average depth is 4 feet deep and it is 10 feet long. If there are 7.48 gallons in a cubic foot, how wide is the pool?

Possible Answers:

133 ft

33 ft

7.48 ft

100 ft

30 ft

Correct answer:

33 ft

Explanation:

There are 7.48 gallons in cubic foot. Set up a ratio:

1 ft³ / 7.48 gallons = x cubic feet / 10,000 gallons

Pool Volume = 10,000 gallons = 10,000 gallons * (1 ft³/ 7.48 gallons) = 1336.9 ft³

Pool Volume = 4ft x 10 ft x WIDTH = 1336.9 cubic feet

Solve for WIDTH:

4 ft x 10 ft x WIDTH = 1336.9 cubic feet

WIDTH = 1336.9 / (4 x 10) = 33.4 ft

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Example Question #11 : Solid Geometry

A cube has a volume of 64cm³. What is the area of one side of the cube?

Possible Answers:

16cm

16cm³

16cm²

4cm

4cm²

Correct answer:

16cm²

Explanation:

The cube has a volume of 64cm³, making the length of one edge 4cm (4 * 4 * 4 = 64).

So the area of one side is 4 * 4 = 16cm²

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Example Question #2 : How To Find The Length Of An Edge Of A Cube

Given that the suface area of a cube is 72, find the length of one of its sides.

Possible Answers:

$\displaystyle 6$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$

$\displaystyle 3$

$3\sqrt{2}$ $\displaystyle 3\sqrt{2}$

$\sqrt{6}$ $\displaystyle \sqrt{6}$

Correct answer:

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$

Explanation:

The standard equation for surface area is

SA=6a^2 $\displaystyle SA=6a^2$

where $\displaystyle a$ denotes side length. Rearrange the equation in terms of $\displaystyle a$ to find the length of a side with the given surface area:

$a=\sqrt{\frac{SA}{6}}=\sqrt{\frac{72}{6}}=\sqrt{12}=2\sqrt{3}$ $\displaystyle a=\sqrt{\frac{SA}{6}}=\sqrt{\frac{72}{6}}=\sqrt{12}=2\sqrt{3}$

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High School Math : Solid Geometry

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : How To Find The Surface Area Of A Polyhedron

Example Question #3 : Other Polyhedrons

Example Question #2 : How To Find The Surface Area Of A Polyhedron

Example Question #21 : Solid Geometry

Example Question #6 : Other Polyhedrons

Example Question #22 : Solid Geometry

Example Question #8 : Other Polyhedrons

Example Question #643 : Geometry

Example Question #11 : Solid Geometry

Example Question #2 : How To Find The Length Of An Edge Of A Cube

All High School Math Resources

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