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SAT II Math I : SAT Subject Test in Math I

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

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6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Advanced Geometry

Find the surface area of a cone with a base diameter of and a height of .

Possible Answers:

$3\pi\sqrt{109} +9\pi$

$3\pi \sqrt{109}$

$12\pi\sqrt{109}$

$30\pi$

$39\pi$

Correct answer:

$3\pi\sqrt{109} +9\pi$

Explanation:

The Surface Area of a cone is:

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

Given the base diameter is 6, the radius of the base is 3. The height is 10. We will substitute these values to find the slant height by using the Pythagorean Theorem.

r^2 + h^2 = s^2

(3)^2 + (10)^2 = s^2

109= s^2

$\sqrt{109} = s$

Substitute slant height and radius into the Surface Area equation.

$SA=\pi (3)(\sqrt{109}) + \pi (3)^2 = 3\pi \sqrt{109} + 9\pi$

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

What is the surface area of a cone with a radius of 4 and a height of 3?

Possible Answers:

$16\pi$

$36\pi$

$25\pi$

$40\pi$

$48\pi$

Correct answer:

$36\pi$

Explanation:

Here we simply need to remember the formula for the surface area of a cone and plug in our values for the radius and height.

$\Pi r^{2} + \Pi r\sqrt{r^{2} + h^{2}}= \Pi\ast 4^{2} + \Pi \ast 4\sqrt{4^{2} + 3^{2}} = 16\Pi + 4\Pi \sqrt{25} = 16\Pi + 20\Pi = 36\Pi$

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

The lateral area is twice as big as the base area of a cone. If the height of the cone is 9, what is the entire surface area (base area plus lateral area)?

Possible Answers:

90π

27π

9π

81π

54π

Correct answer:

81π

Explanation:

Lateral Area = LA = π(r)(l) where r = radius of the base and l = slant height

LA = 2B

π(r)(l) = 2π(r²)

rl = 2r²

l = 2r

Cone

From the diagram, we can see that r² + h² = l². Since h = 9 and l = 2r, some substitution yields

r² + 9² = (2r)²

r² + 81 = 4r²

81 = 3r²

27 = r²

B = π(r²) = 27π

LA = 2B = 2(27π) = 54π

SA = B + LA = 81π

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

A right cone has a radius of 4R and a height of 3R. What is the ratio of the total surface area of the cone to the surface area of just the base?

Possible Answers:

$\frac{3}{2}$

$\frac{9}{5}$

$\frac{9}{4}$

$\frac{7}{6}$

Correct answer:

$\frac{9}{4}$

Explanation:

We need to find total surface area of the cone and the area of the base.

The area of the base of a cone is equal to the area of a circle. The formula for the area of a circle is given below:

$A=\pi r^2$ , where r is the length of the radius.

In the case of this cone, the radius is equal to 4R, so we must replace r with 4R.

$A=\pi(4R)^2=\pi(4)^2(R)^2=16\pi R^2$

To find the total area of the cone, we need the area of the base and the lateral surface area of the cone. The lateral surface area (LA) of a cone is given by the following formula:

$LA=\frac{1}{2}(2\pi r)(l)$ , where r is the radius and l is the slant height.

We know that r = 4R. What we need now is the slant height, which is the distance from the edge of the base of the cone to the tip.

In order to find the slant height, we need to construct a right triangle with the legs equal to the height and the radius of the cone. The slant height will be the hypotenuse of this triangle. We can use the Pythagorean Theorem to find an expression for l. According to the Pythagorean Theorem, the sum of the squares of the legs (which are 4R and 3R in this case) is equal to the square of the hypotenuse (which is the slant height). According to the Pythagorean Theorem, we can write the following equation:

(4R)^2+(3R)^2=l^2

16R^2+9R^2=l^2

25R^2=l^2

l=5R

Let's go back to the formula for the lateral surface area (LA).

$LA=\frac{1}{2}(2\pi r)(l)$

$LA=\frac{1}{2}(2\pi(4R))(5R)$

$LA=20\pi R^2$

To find the total surface area (TA), we must add the lateral area and the area of the base.

$TA=20\pi R^2+16\pi R^2=36\pi R^2$

The problem requires us to find the ratio of the total surface area to the area of the base. This means we must find the following ratio:

$\frac{36\pi R^2}{16\pi R^2}$

We can cancel $\pi R^2$ , which leaves us with 36/16.

Simplifying 36/16 gives 9/4.

The answer is 9/4.

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Example Question #12 : Cones

What is the surface area of a cone with a radius of 6 in and a height of 8 in?

Possible Answers:

36π in²

60π in²

66π in²

112π in²

96π in²

Correct answer:

96π in²

Explanation:

Find the slant height of the cone using the Pythagorean theorem: r² + h² = s² resulting in 6² + 8² = s² leading to s² = 100 or s = 10 in

SA = πrs + πr² = π(6)(10) + π(6)² = 60π + 36π = 96π in²

60π in² is the area of the cone without the base.

36π in² is the area of the base only.

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

Use the following formula to answer the question.

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

The slant height of a right circular cone is 6cm . The radius is 2cm , and the height is 4cm . Determine the surface area of the cone.

Possible Answers:

$12\pi cm^2$

$16\pi cm^2$

$18\pi cm^2$

$24\pi cm^2$

$20\pi cm^2$

Correct answer:

$16\pi cm^2$

Explanation:

Notice that the height of the cone is not needed to answer this question and is simply extraneous information. We are told that the radius is 2cm , and the slant height is 6cm .

First plug these numbers into the equation provided.

$SA=\pi r^2+\pi rl=\pi (2)^2+\pi (2)(6)=4\pi cm^2+12\pi cm^2$

Then simplify by combining like terms.

$4\pi cm^2+12\pi cm^2=16\pi cm^2$

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Example Question #11 : Surface Area

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

Possible Answers:

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{20}$

$\frac{ \pi L^{2}}{5}$

$\frac{6\pi L^{2}}{25}$

$\frac{11\pi L^{2}}{100}$

Correct answer:

$\frac{11\pi L^{2}}{100}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

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

The radius of the base of a cone is ; its slant height is two-thirds of the diameter of that base. Give its surface area in terms of .

Possible Answers:

$\frac{5\pi r ^{2}}{3}$

$\frac{4\pi r ^{2}}{3}$

$\frac{7\pi r ^{2}}{3}$

$\frac{\left (6- \sqrt{7} \right )\pi r ^{2}}{3}$

$\frac{\left (3- \sqrt{5} \right )\pi r ^{2}}{3}$

Correct answer:

$\frac{7\pi r ^{2}}{3}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The diameter of the base is twice radius , or , and its slant height is two-thirds of this diameter, which is $\frac{2}{3}\cdot 2r = \frac{4}{3}r$ . Substitute this for in the formula:

$A = \pi r^{2} + \pi r \cdot \frac{4}{3}r$

$A = \pi r^{2} + \frac{4}{3} \pi r ^{2}$

$A = \frac{7}{3} \pi r ^{2} = \frac{7\pi r ^{2}}{3}$

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

The radius of the base of a cone is ; its height is twice of the diameter of that base. Give its surface area in terms of .

Possible Answers:

$3 \pi r^{2}$

$5 \pi r^{2}$

$\left ( 1+ \sqrt{5} \right ) \pi r^{2}$

$\left ( 1+2 \sqrt{5} \right ) \pi r^{2}$

$\left ( 1+ \sqrt{17} \right ) \pi r^{2}$

Correct answer:

$\left ( 1+ \sqrt{17} \right ) \pi r^{2}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The base has radius and diameter . The height is twice the diamter, which is . Its slant height can be calculated using the Pythagorean Theorem:

$l = \sqrt{r^{2}+ h^{2}}$

$l = \sqrt{r^{2}+ (4r)^{2}}$

$l = \sqrt{r^{2}+ 16r^{2}}$

$l = \sqrt{17r^{2}}$

$l = r \sqrt{17 }$

Substitute $r \sqrt{17 }$ for in the surface area formula:

$A = \pi r^{2} + \pi r \cdot r \sqrt{17}$

$A = \pi r^{2} + \pi r^{2} \sqrt{17}$

$A = \left ( 1+ \sqrt{17} \right ) \pi r^{2}$

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Example Question #122 : Geometry

The height of a cone is ; the diameter of its base is twice the height. Give its surface area in terms of .

Possible Answers:

$10 \pi h^{2}$

$\left (2 +2 \sqrt{2} \right ) \pi h^{2}$

$5 \pi h^{2}$

$4 \pi h^{2}$

$\left (1 + \sqrt{2} \right ) \pi h^{2}$

Correct answer:

$\left (1 + \sqrt{2} \right ) \pi h^{2}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The diameter of the base is twice the height, which is ; the radius is half this, which is .

The slant height can be calculated using the Pythagorean Theorem:

$l = \sqrt{r^{2}+ h^{2}}$

$l = \sqrt{h^{2}+ h^{2}}$

$l = \sqrt{2h^{2}}$

$l =h \sqrt{2}$

Substitute $h \sqrt{2}$ for and for in the surface area formula:

$A = \pi h^{2} + \pi h \cdot h \sqrt{2}$

$A = \pi h^{2} + \pi h ^{2} \sqrt{2}$

$A = \left (1 + \sqrt{2} \right ) \pi h^{2}$

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SAT II Math I : SAT Subject Test in Math I

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

All SAT II Math I Resources

Example Questions

Example Question #4 : Advanced Geometry

Example Question #1 : 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 Cone

Example Question #12 : Cones

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

Example Question #11 : Surface Area

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

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

Example Question #122 : Geometry

All SAT II Math I Resources

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