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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

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8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : How To Find The Surface Area Of A Cone

Find the surface area of a cone that has a radius of 12 and a slant height of 15.

Possible Answers:

$180\pi$

$324\pi$

$405\pi$

$144\pi$

$225\pi$

Correct answer:

$324\pi$

Explanation:

The standard equation to find the surface area of a cone is

$SA=\pi rs+\pi r^2$

where denotes the slant height of the cone, and denotes the radius.

Plug in the given values for and to find the answer:

$SA=\pi (12)(15)+\pi (12)^2=180\pi +144\pi =324\pi$

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Example Question #4 : How To Find The Surface Area Of A Cone

Find the surface area of the following cone.

Cone

Possible Answers:

$100 \pi m^2$

$85 \pi m^2$

$90 \pi m^2$

$95 \pi m^2$

$80 \pi m^2$

Correct answer:

$90 \pi m^2$

Explanation:

The formula for the surface area of a cone is:

$SA = A_{base} + A_{lateral}$

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

where is the radius of the cone and is the slant height of the cone.

Plugging in our values, we get:

$SA = \pi (5m)^2 + \pi (5m) (13m)$

$SA = 25 \pi m^2 + 65 \pi m^2 = 90 \pi m^2$

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Example Question #5 : How To Find The Surface Area Of A Cone

Find the surface area of the following cone.

Cone

Possible Answers:

$116 \pi m^2$

$86 \pi m^2$

$96 \pi m^2$

$106 \pi m^2$

$76 \pi m^2$

Correct answer:

$96 \pi m^2$

Explanation:

The formula for the surface area of a cone is:

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

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

$SA = (\pi r^2) + ( \pi rh_s)$

Use the Pythagorean Theorem to find the length of the radius:

A^2 + B^2 = C^2

A^2 + (8m)^2 = (10m)^2

A = 6m

Plugging in our values, we get:

$SA = \pi (6m)^2 + \pi (6m)(10m)$

$SA = 96 \pi m^2$

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

Find the surface area of the following half cone.

Half_cone

Possible Answers:

$120 \pi m^2 + 120 m^2$

$88 \pi m^2$

$88 \pi m^2 + 88 m^2$

$120 \pi m^2$

$100 \pi m^2 + 120 m^2$

Correct answer:

$100 \pi m^2 + 120 m^2$

Explanation:

The formula for the surface area of the half cone is:

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

$SA=\left(\frac{\pi r^2}{2}\right)+\left(\frac{1}{2}\cdot \frac{2\pi r h_s}{2}\right)+rh$

Where is the radius, h_s is the slant height, and is the height of the cone.

Use the Pythagorean Theorem to find the height of the cone:

A^2+B^2 = C^2

A^2+(8m)^2=(17m)^2

A=15m

Plugging in our values, we get:

$SA=\left(\frac{\pi (8m)^2}{2}\right)+\left( \frac{\pi (8m) (17m)}{2}\right)+(8m)(15m)$

$SA=32 \pi m^2 + 68 \pi m^2 + 120m^2$

$SA = 100 \pi m^2 + 120 m^2$

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

Find the surface area of the following half-cylinder.

Half_cylinder

Possible Answers:

$90 \pi m^2$

$150+100 \pi m^2$

$50+125 \pi m^2$

$110 \pi m^2$

$75\pi m^2$

Correct answer:

$150+100 \pi m^2$

Explanation:

The formula for the surface area of a half-cylinder must include one-half of the surface area of a cylinder, which would be:

$SA = \frac{1}{2} (lateral\ area + 2 \cdot (base))$

We also need to add the area of the new rectangular face that is created by cutting the cylinder in half. The area of this rectangle would be:

$A=l\times w$

where the length of the rectangle is the same as the height of the half-cylinder, and the width of the rectangle is the same as the diameter of the base of the half-cylinder. So we can rewrite the area of the rectangle as:

$A=h\p\times d$

Now we can combine the two area formulas to find the total surface area of the half-cylinder:

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

$SA = \pi r (h) + \pi r^2+hd$

where is the radius of the base and is the length of the height, and is the diameter of the base.

Plugging in our values, we get:

$SA = \pi (5m) (15m) + \pi (5m)^2+15\cdot 10$

$SA = 75 \pi m^2 + 25 \pi m^2+150 = 150+100 \pi m^2$

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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:

$98 \pi m^2$

$108 \pi m^2$

$118 \pi m^2$

$88 \pi m^2$

$128 \pi m^2$

Correct answer:

$108 \pi m^2$

Explanation:

The formula for the surface area of the polyhedron is:

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

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

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

Where is the radius of the cone, is the slant height of the cone, and is the radius of the sphere

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

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

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

Plugging in our values, we get:

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

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

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

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Example Question #1 : Other Polyhedrons

Find the surface area of the following polyhedron.

Dome

Possible Answers:

$101 \pi m^2$

$71 \pi m^2$

$61 \pi m^2$

$91 \pi m^2$

$81 \pi m^2$

Correct answer:

$81 \pi m^2$

Explanation:

The formula for the surface area of the polyhedron is:

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

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

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

where is the radius of the cone, h_s is the slant height of the cone, is the radius of the cylinder, and is the height of the cylinder.

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

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

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

Plugging in our values, we get:

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

$SA = 81 \pi m^2$

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Example Question #1 : Other Polyhedrons

Find the surface area of the following polyhedron.

Ice_cream_cone

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The formula for the surface area of a polyhedron is:

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

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

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

where is the radius of the polyhedron and h_s is the slant height of the cone.

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

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

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

Plugging in our values, we get:

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

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

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Example Question #2 : Other Polyhedrons

Find the volume of the following half cylinder.

Half_cylinder

Possible Answers:

$150.5 \pi m^3$

$187.5 \pi m^3$

$207.5 \pi m^3$

$200 \pi m^3$

$157.5 \pi m^3$

Correct answer:

$187.5 \pi m^3$

Explanation:

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

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

where is the radius of the base and is the length of the height.

Plugging in our values, we get:

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

$V = 187.5 \pi m^3$

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

Find the volume of the following polyhedron.

Ice_cream_cone

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The formula for the volume of the polyhedron is:

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

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

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

Where is the radius of the cone, is the height of the cone, and is the radius of the sphere.

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

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

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

Plugging in our values, we get:

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

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

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

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : How To Find The Surface Area Of A Cone

Example Question #4 : How To Find The Surface Area Of A Cone

Example Question #5 : How To Find The Surface Area Of A Cone

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

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

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

Example Question #1 : Other Polyhedrons

Example Question #1 : Other Polyhedrons

Example Question #2 : Other Polyhedrons

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

All High School Math Resources

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