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SAT II Math I : Surface Area

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

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Example Questions

Example Question #21 : 3 Dimensional Geometry

A circular swimming pool has diameter 32 meters and depth meters throughout. Which of the following expressions gives the total area of the inside of the pool, in square meters?

Possible Answers:

None of the other responses is correct.

$256\pi + 16 \pi t$

$512\pi + 32 \pi t$

$256\pi + 32 \pi t$

$512\pi + 16 \pi t$

Correct answer:

$256\pi + 32 \pi t$

Explanation:

The bottom of the pool is a circle with diameter 32, and, subsequently, radius half this, or 16; its area is

$B = \pi r^{2} = \pi \times 16^{2} = 256 \pi$

The side of the pool is the lateral surface of a cylinder with radius 16 and height ; the area of this is

$LA = 2 \pi r h = 2 \pi \cdot 16 \cdot t = 32 \pi t$

The area of the inside of the pool is the sum of these two, or

$A = 256 \pi + 32 \pi t$

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Example Question #1 : Surface Area

A circular swimming pool at an apartment complex has diameter 50 feet and depth six feet throughout.

The apartment manager needs to get the interior of the swimming pool painted. The paint she wants to use covers 350 square feet per gallon. How many one-gallon cans of paint will she need to purchase?

Possible Answers:

The correct answer is not given among the other responses.

Correct answer:

Explanation:

The pool can be seen as a cylinder with depth (or height) six feet and a base with diameter 50 feet - and radius half this, or 25 feet.

The bottom of the pool - the base of the cylinder - is a circle with radius 25 feet, so its area is

$B = \pi r ^{2} = 3.14 \cdot 25 \cdot 25 = 1,962.5$ square feet.

Its side - the lateral face of the cylinder - has area

$LA = 2 \pi r h = 2 \cdot 3.14 \cdot 25 \cdot 6 = 942$ square feet.

Their sum - the total area to be painted - is 1,962.5 + 942 = 2,904.5 square feet. Since one gallon of paint covers 350 square feet, divide:

$2,904.5 \div 350 \approx 8.3$

Eight cans of paint and part of a ninth will be required, so the correct response is nine.

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Example Question #2 : Surface Area

Example cylinder

Figure not drawn to scale.

Find the surface area of the cylinder above.

Possible Answers:

56.55 in²

73.44 in²

87.25 in²

122.14 in²

94.25 in²

Correct answer:

94.25 in²

Explanation:

In order to find the surface area of a cylinder, you need to find the surface areas of the circles that are the top and bottom of the cylinder (2 x pi x radius²) and add it to the surface are of the rectangle that is the side of the cylinder (diameter x height).

$\small \small \small \small Surface Area=(2\pi r{^2})+(\pi2r\times h)$

$\small \small (2\pi\times 3{^2})+(6\pi\times 2)$

$\small \small (6.28\times 9)+(18.84\times 2)$

$\small 56.54 + 37.70=94.25$

The surface area of the cylinder is 94.25 in²

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Example Question #1 : Surface Area

Example images

Figure not drawn to scale

What is the surface area of the sphere above?

Possible Answers:

712.12 yd²

512.63 yd²

815.44 yd²

615.75 yd²

615.75 yd³

Correct answer:

615.75 yd²

Explanation:

In order to find the surface area of a sphere, you must use the equation below:

$\small Surface Area=4\pi r{^{2}}$

$\small 4\pi(7{^{2}})=4\pi (49)=196\pi=615.75$

The surface area of the sphere is 615.75 yd²

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Example Question #1 : Advanced Geometry

What is the surface area of the following cone?

Possible Answers:

$\pi$ m^2

Correct answer:

$\pi$ m^2

Explanation:

The formula for the surface area of a cone is:

A = $\pi$ $\pi$ ,

where represents the radius of the cone base and represents the slant height of the cone.

Plugging in our values, we get:

A = $\pi$ $\pi$

A = $\pi$ m^2

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Example Question #2 : Advanced Geometry

The surface area of cone $\small E$ is $\small 70\pi$ . If the radius of the base of the cone is , what is the height of the cone?

Possible Answers:

$\small 2\sqrt{5}$

$\small 3$

$\small 9$

$\small \small 2\sqrt{14}$

$\small 45$

Correct answer:

$\small \small 2\sqrt{14}$

Explanation:

To figure out $\small h$ , we must use the equation for the surface area of a cone, $\small A=\pi r^2+\pi rs$ , where $\small r$ is the radius of the base of the cone and $\small s$ is the length of the diagonal from the tip of the cone to any point on the base's circumference. We therefore first need to solve for $\small s$ by plugging what we know into the equation:

$\small 70\pi=(5^2)\pi +5s\pi=25\pi +5s\pi$

This equation can be reduced to:

$\small 45\pi=5s\pi$

$\small 9=s$

For a normal right angle cone, $\small s$ represents the line from the tip of the cone running along the outside of the cone to a point on the base's circumference. This line represents the hypotenuse of the right triangle formed by the radius and height of the cone. We can therefore solve for $\small h$ using the Pythagorean theorem:

$\small s^2 = r^2+h^2$

$\small h=\sqrt{s^2-r^2}$

Our $\small h$ is therefore:

$\small h=\sqrt{9^2-5^2}=\sqrt{81-25}=\sqrt{56}=2\sqrt{14}$

The height of cone $\small E$ is therefore $\small 2\sqrt{14}$

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Example Question #1 : Solid Geometry

A circle of radius five is cut into two pieces, $30\%$ and $70\%$ . The larger section is thrown away. The smaller section is curled until the two straight edges meet, and a bottom is made for the cone.

What is the area of the bottom?

Possible Answers:

$2\pi$

$25\pi$

$3\pi$

$\frac{9\pi }{4}$

Correct answer:

$\frac{9\pi }{4}$

Explanation:

When the smaller portion of the circle is curled in, it will make the top of a cone. The circumfrence of the circle on the bottom is $2\pi r$ (where r is the radius of the circle on the bottom). The circumference of the bottom is also $30\%$ of the circumfrence of the original larger circle, which is $.3C=.3(2\pi R)=.3(10\pi)=3\pi$ (where R is the radius of the original, larger circle)

Therefore we use the circumference formula to solve for our new r:

$2\pi r=3\pi, r=\frac{3}{2}$

Substituting this value into the area formula, the area of the small circle becomes:

$A=\pi r^2=\frac{9\pi}{4}$

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Example Question #3 : Advanced Geometry

A cone has a bottom area of $9\pi$ and a height of , what is the surface area of the cone?

Possible Answers:

$20\pi$

$24\pi$

$12\pi$

Correct answer:

$24\pi$

Explanation:

The area of the bottom of the cone yields the radius,

$A=\pi r^2, r=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{9\pi}{\pi}}=3$

The height of the cone is , so the Pythagorean Theorem will give the slant height, $l=\sqrt{3^2+4^2}=5$

The area of the side of the cone is $A=\pi rl=\pi \cdot 3 \cdot 5=15\pi$ and adding that to the $9\pi$ given as the area of the circle, the surface area comes to $24\pi$

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Example Question #1 : Solid Geometry

If the surface area of a right angle cone $\small A$ is $\small 48\pi$ , and the distance from the tip of the cone to a point on the edge of the cone's base is $\small 8$ , what is the cone's radius?

Possible Answers:

$16\pi$

Correct answer:

Explanation:

Solving this problem is going to take knowledge of Algebra, Geometry, and the equation for the surface area of a cone: $\small \small A=\pi r^2 + \pi r s$ , where $\small r$ is the radius of the cone's base and $\small s$ is the distance from the tip of the cone to a point along the edge of the cone's base. First, let's substitute what we know in this equation:

$\small \small 48\pi=\pi r^2 + \pi r (8)=\pi r^2 + 8\pi r$

We can divide out $\small \small \pi$ from every term in the equation to obtain:

$\small \small \small 48=r^2+8r\Rightarrow0=r^2+8r-48$

We see this equation has taken the form of a quadratic expression, so to solve for $\small r$ we need to find the zeroes by factoring. We therefore need to find factors of $\small \small -48$ that when added equal $\small \small 8$ . In this case, $\small -4$ and $\small \small 12$ :

$\small \small r^2+8r-48=(r+12)(r-4)$

This gives us solutions of $\small \small r=-12$ and $\small r=4$ . Since $\small r$ represents the radius of the cone and the radius must be positive, we know that $\small r=4$ is our only possible answer, and therefore the radius of the cone is $\small 4$ .

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Example Question #1 : Advanced Geometry

For a right circular cone , the radius is and the height of the cone is . What is the surface area of the cone in terms of $\pi$ ?

Possible Answers:

$30\pi$

$\small 100\pi+10\pi\sqrt{109}$

$130\pi$

$\small 209\pi$

$\small 200\pi$

Correct answer:

$\small 100\pi+10\pi\sqrt{109}$

Explanation:

To solve this problem, we will need to use the formula for finding the surface area of a cone, $\small A=\pi r^2+\pi rs$ , where $\small s$ is the length of the diagonal from the circle edge of the cone to the top. Since we are not given s, we must find it by using Pythagorean's Theorem:

$\small s=\sqrt{r^2+h^2}=\sqrt{10^2+3^2}=\sqrt{109}$ .

$\small 109$ is a prime number, so we cannot factor the radical any further. Therefore, our equation for our surface area of $\small C$ becomes:

$\small A=\pi(10^2)+\pi(10)\sqrt{109}=100\pi + 10\pi\sqrt{109}$ , which is our final answer.

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SAT II Math I : Surface Area

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

All SAT II Math I Resources

Example Questions

Example Question #21 : 3 Dimensional Geometry

Example Question #1 : Surface Area

Example Question #2 : Surface Area

The surface area of the cylinder is 94.25 in²

Example Question #1 : Surface Area

What is the surface area of the sphere above?

The surface area of the sphere is 615.75 yd²

Example Question #1 : Advanced Geometry

Example Question #2 : Advanced Geometry

Example Question #1 : Solid Geometry

Example Question #3 : Advanced Geometry

Example Question #1 : Solid Geometry

Example Question #1 : Advanced Geometry

All SAT II Math I Resources

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HIGH SCHOOL

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K-8

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SAT II Math I : Surface Area

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

All SAT II Math I Resources

Example Questions

Example Question #21 : 3 Dimensional Geometry

Example Question #1 : Surface Area

Example Question #2 : Surface Area

The surface area of the cylinder is 94.25 in2

Example Question #1 : Surface Area

What is the surface area of the sphere above?

The surface area of the sphere is 615.75 yd2

Example Question #1 : Advanced Geometry

Example Question #2 : Advanced Geometry

Example Question #1 : Solid Geometry

Example Question #3 : Advanced Geometry

Example Question #1 : Solid Geometry

Example Question #1 : Advanced Geometry

All SAT II Math I Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

The surface area of the cylinder is 94.25 in²

The surface area of the sphere is 615.75 yd²