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ISEE Upper Level Quantitative : Geometry

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

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All ISEE Upper Level Quantitative Resources

8 Diagnostic Tests 234 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #5 : How To Find The Volume Of A Cylinder

What is the volume of a cylinder with a height of in. and a radius of 2.5 in?

Possible Answers:

$87.5\pi$ in^3

$82.5\pi$ in^3

$42.25\pi$ in^3

$70\pi$ in^3

$35\pi$ in^3

Correct answer:

$87.5\pi$ in^3

Explanation:

This is a rather direct question. Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$

For our values this is:

$A = \pi * 2.5^2 * 14=87.5\pi$

This is the volume of the cylinder.

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Example Question #351 : Geometry

What is the radius of a cylinder with a volume of $562.5\pi$ in^3 and a height of ?

Possible Answers:

$56.25\pi$

56.25

$15\pi$

7.5

Correct answer:

7.5

Explanation:

Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$

For our values this is:

$562.5\pi = \pi * r^2 * 10$

Solve for :

56.25=r^2

Using a calculator to calculate $\sqrt{56.25}$ , you will see that r = 7.5

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Example Question #1 : Cylinders

What is the surface area of a cylinder of height in., with a radius of in?

Possible Answers:

$136\pi$ in^2

$152\pi$ in^2

$120\pi$ in^2

$122\pi$ in^2

$240\pi$ in^2

Correct answer:

$152\pi$ in^2

Explanation:

Recall that to find the surface area of a cylinder, you need to find the surface area of its two bases and then the surface area of its "outer face." The first two are very easy since they are circles. The equation for one base is:

$A = \pi * r^2$

For our problem, this is:

$\pi * 4^2 = \pi * 16$

You need to double this for the two bases:

$2 * 16 * \pi = 32 * \pi$

The area of the "outer face" is a little bit trickier, but it is not impossible. It is actually a rectangle that has the height of the cylinder and a width equal to the circumference of the base; therefore, it is:

$2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 4 * 15 = 120 * \pi$

Therefore, the total surface area is:

$32\pi + 120\pi = 152\pi$

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Example Question #2 : Cylinders

What is the surface area of a cylinder having a base of radius in and a height of in?

Possible Answers:

$212\pi$ in^2

$308\pi$ in^2

$99\pi$ in^2

$363\pi$ in^2

$33\pi$ in^2

Correct answer:

$308\pi$ in^2

Explanation:

$A = \pi * r^2$

For our problem, this is:

$\pi * 11^2 = \pi * 121$

You need to double this for the two bases:

$2 * 121 * \pi = 242 * \pi$

$2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 11* 3 = 66 * \pi$

Therefore, the total surface area is:

$242 * \pi + 66 * \pi = 308\pi$

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Example Question #3 : Cylinders

What is the surface area of a cylinder with a height of in. and a diameter of in?

Possible Answers:

$22\pi$ in^2

$57\pi$ in^2

$28\pi$ in^2

$36\pi$ in^2

$14\pi$ in^2

Correct answer:

$36\pi$ in^2

Explanation:

$A = \pi * r^2$

For our problem, this is:

$\pi * 2^2 = \pi * 4$

You need to double this for the two bases:

$2 * 4 * \pi = 8 * \pi$

$2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 2 * 7 = 28 * \pi$

Therefore, the total surface area is:

$28\pi + 8\pi = 36\pi$

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Example Question #51 : Solid Geometry

The volume of a cylinder with height of is $275\pi$ in^3 . What is its surface area?

Possible Answers:

$55\pi$ in^2

$66\pi$ in^2

$82\pi$ in^2

$135\pi$ in^2

$160\pi$ in^2

Correct answer:

$160\pi$ in^2

Explanation:

To begin, we must solve for the radius of this cylinder. Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$

For our values this is:

$275\pi = \pi * r^2 * 11$

Solving for , we get:

r^2 = 25

Hence, r = 5

Now, recall that to find the surface area of a cylinder, you need to find the surface area of its two bases and then the surface area of its "outer face." The first two are very easy since they are circles. The equation for one base is:

$A = \pi * r^2$

For our problem, this is:

$\pi * 5^2 = \pi * 25$

You need to double this for the two bases:

$2 * 25 * \pi = 50 * \pi$

$2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 5 * 11 = 110 * \pi$

Therefore, the total surface area is:

$50\pi + 110\pi = 160\pi$

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

What is the surface area of a cylinder of height in, with a radius of in?

Possible Answers:

$136\pi$ in^2

$122\pi$ in^2

$152\pi$ in^2

$240\pi$ in^2

$120\pi$ in^2

Correct answer:

$152\pi$ in^2

Explanation:

$A = \pi * r^2$

For our problem, this is:

$\pi * 4^2 = \pi * 16$

You need to double this for the two bases:

$2 * 16 * \pi = 32 * \pi$

$2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 4 * 15 = 120 * \pi$

Therefore, the total surface area is:

$32\pi + 120\pi = 152\pi$

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Geometry

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #5 : How To Find The Volume Of A Cylinder

Example Question #351 : Geometry

Example Question #1 : Cylinders

Example Question #2 : Cylinders

Example Question #3 : Cylinders

Example Question #51 : Solid Geometry

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

All ISEE Upper Level Quantitative Resources

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