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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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Example Questions

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 #1 : Cones

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:

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

$\small 2\sqrt{5}$

$\small 45$

$\small 3$

$\small 9$

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 #2 : Advanced 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:

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

$2\pi$

$25\pi$

$3\pi$

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:

$24\pi$

$20\pi$

$12\pi$

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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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 #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 #1 : Cones

Example Question #2 : Advanced Geometry

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

All SAT II Math I Resources

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HIGH SCHOOL

GRADUATE SCHOOL

K-8

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math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

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 #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 #1 : Cones

Example Question #2 : Advanced Geometry

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

All SAT II Math I Resources

Report an issue with this question

DMCA Complaint

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Complaint Details

The surface area of the sphere is 615.75 yd²