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SAT II Math I : Geometry

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

Example Questions

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:

$24\pi$

$20\pi$

$12\pi$

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 #4 : Advanced 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 #5 : 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:

$\small 209\pi$

$30\pi$

$\small 200\pi$

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

$130\pi$

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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Example Question #6 : Advanced Geometry

If the surface area of cone $\small D$ is $\small 55\pi$ , and the distance $\small s$ between the cone's tip and a point on the cone's circular base is $\small 6$ , what is the radius $\small r$ of the cone?

Possible Answers:

$\small 5$

$\small \small 10$

$\small 3$

$\small 10\pi$

$\small 6$

Correct answer:

$\small 5$

Explanation:

To find out the radius, we must use our knowledge of the formula for the surface area of a cone: $\small A= \pi r^2 + \pi rs$ , where $\small r$ is the radius of the cone and $\small s$ is the distance from the tip of the cone to any point along the circumference of the cone's base. We can plug in what we already know into the above equation:

$\small 55\pi=\pi r^2+6\pi r$

We can divide out $\small \pi$ from each term to obtain:

$\small 55=r^2+6r\Rightarrow0=r^2+6r-55$

We now can recognize that the above is a quadratic expression, so to solve for $\small r$ we can find the zeroes of the equation by factoring. We need two numbers which will multiply to $\small -55$ but will add to $\small 6$ (in this case $\small 11$ and $\small -5$ ). Therefore, we can factor the above to the following:

$\small r^2+6r-55=(r+11)(r-5)$ .

Our two solutions are therefore $\small r=-11$ and $\small r=5$ . Since $\small r$ represents the radius of the base of the cone, it must be positive, and that leaves $\small 5$ as our one and only answer.

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Example Question #1 : Advanced Geometry

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

Possible Answers:

$96\pi$

$39\pi$

$30\pi$

$72\pi$

$90\pi$

Correct answer:

$39\pi$

Explanation:

The Surface Area of a cone is:

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

Given the base diameter is 6, the radius will be 3. The given slant height is 10.

Substitute the radius and slant height into the equation to find surface area.

$SA=\pi (3)(10) +\pi (3)^2 = 30\pi + 9\pi = 39\pi$

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Example Question #2 : Advanced Geometry

Find the surface area of a cone with a base area of $4\pi$ and a slant height of .

Possible Answers:

$8\sqrt2\pi +4\pi$

$\frac{16\sqrt2}{3} \pi$

$16\pi$

$24\pi$

$8\pi$

Correct answer:

$16\pi$

Explanation:

The surface area of a cone is:

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

Given the base area is $4\pi$ , the base of the cone resembles a circle. Using the base area, it is necessary to find the radius.

$A = \pi r^2$

$4\pi = \pi r^2$

4=r^2

2=r

Since radius of the base is 2, and slant height is 6, substitute these into the surface area equation.

$SA= \pi (2)(6) + \pi (2)^2=12\pi+4\pi=16\pi$

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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:

$\frac{9}{4}$

$\frac{7}{6}$

$\frac{9}{5}$

$\frac{3}{2}$

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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SAT II Math I : Geometry

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

All SAT II Math I Resources

Example Questions

Example Question #3 : Advanced Geometry

Example Question #4 : Advanced Geometry

Example Question #5 : Advanced Geometry

Example Question #6 : Advanced Geometry

Example Question #1 : Advanced Geometry

Example Question #2 : Advanced Geometry

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

All SAT II Math I Resources

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