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SAT II Math I : 3-Dimensional Geometry

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

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Example Questions

Example Question #21 : Advanced Geometry

The circumference of the base of a cone is 100; the height of the cone is equal to the diameter of the base. Give the surface area of the cone (nearest whole number).

Possible Answers:

2,575

2,385

1,589

1,779

8,443

Correct answer:

2,575

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 the circumference divided by $\pi$ , which is

$d = C \div (2\pi ) \approx 100 \div 3.14 \approx 31.83$

This is also the height .

The radius is half this, or

$r = d \div 2 \approx 31.83 \div 2 \approx 15.92$

The slant height can be found by way of the Pythagorean Theorem:

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

$l \approx \sqrt{15.92^{2}+ 31.83 ^{2}}$

$l \approx \sqrt{253.30+ 1,013.21}$

$l \approx \sqrt{1266.51}$

$l \approx 35.60$

Substitute in the surface area formula:

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

$A \approx 3.14 \cdot 15.92 ^{2} + 3.14 \cdot 15.92 \cdot 35.60$

$A \approx 795.8+ 1,780.1$

$A \approx 2,575$

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Example Question #51 : 3 Dimensional Geometry

If a cone were unfurled into a 2-dimensional figure. The lateral area of the cone would look most like which figure?

Possible Answers:

Triangle

Circle

Sector of a Circle

Rectangle

Correct answer:

Sector of a Circle

Explanation:

When creating a net image of a 3D figure - one imagines it is made of paper and is unfurled into its' 2D form. The lateral portion of the cone cone would be unfurled into the image of a Sector of a Circle. To include the full surface area of the cone a circle is included to form the base of the cone as in the figure below. The lateral area portion is the top part of the figure below.

Cone net

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Example Question #52 : 3 Dimensional Geometry

Find the surface area of a cube with side length of 6in.

Possible Answers:

24cm^2

18cm^2

250cm^2

216cm^2

36cm^2

Correct answer:

216cm^2

Explanation:

A cube is made up of 6 identical sides. Find the area of one side and multiplying it by 6 will result in the surface area of a cube.

S=6a^2

a is the side length, in this case 6in.

S=6(6cm)^2=6*36cm^2=216cm^2

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Example Question #27 : Surface Area

What is the surface area of a cylinder with height 10 and radius 7?

Possible Answers:

$238\pi$

$163\pi$

$140\pi$

$300\pi$

$98\pi$

Correct answer:

$238\pi$

Explanation:

The formula of the surface area of a cylinder is:

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

In this problem h is 10 and r is 7. Substituting in those values gives us:

$SA=2\pi 7*10+2\pi 7^2$

Simplifying and combining the like terms results in:

$SA=140\pi+98\pi=238\pi$

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Example Question #22 : Surface Area

Figure not drawn to scale.

Box example

Find the surface area of the box above.

Possible Answers:

Surface area: 82 yd²

Surface area: 84 yd²

Surface area: 82 yd³

Surface area: 84 yd³

Surface area: 48 yd²

Correct answer:

Surface area: 82 yd²

Explanation:

Box example

You can find the surface area of a box by following the equation below:

$\small \small \small Surface Area = 2(length\times height) + 2(length\times depth) + 2(height\times depth)$

$\small \small \small 2(7\times 3) + 2(7\times 2) + 2(3\times 2)=$

$\small 2(21) + 2(14) + 2(6)=$

$\small \small 42 + 28 + 12= 82$

The surface area of the box is 82yd²(remember that area measurements are square units NOT cubic units)

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Example Question #29 : Surface Area

Box 3

The shaded face of the provided rectangular prism is a square. If the surface area of the prism is , give in terms of .

Possible Answers:

$x = \frac{-25+\sqrt{4A+625}}{4}$

$x = \frac{A}{625}$

$x=\frac{ A - 1,250}{100}$

$x = \frac{-50+\sqrt{2A+2,500}}{2}$

$x=\frac{ A -2,500}{50}$

Correct answer:

$x=\frac{ A - 1,250}{100}$

Explanation:

Since the top face of the prism is a square, the common sidelength - and the missing dimension - is 25.

The surface area of a rectangular prism with length , width , and height can be found using the formula

2LW + 2 WH + 2LH = A .

Setting H= x, L=W= 25 , and solving for :

$2 \cdot 25 \cdot 25 + 2 \cdot 25 \cdot x + 2 \cdot 25 \cdot x = A$

1,250 + 50x + 50 x = A

100 x + 1,250 = A

100 x + 1,250 - 1,250 = A - 1,250

100 x = A - 1,250

$\frac{100 x }{100}=\frac{ A - 1,250}{100}$

$x=\frac{ A - 1,250}{100}$

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Example Question #1 : Faces, Face Area, And Vertices

How many vertices does a polyhedron with twenty faces and thirty edges have?

Possible Answers:

Correct answer:

Explanation:

By Euler's Formula, the relationship between the number of vertices , the number of faces , and the number of edges of a polyhedron is

V + F - E = 2

Set F = 20 and E = 30 , and solve for :

V + 20 - 30 = 2

V -10 = 2

V = 12

The polyhedron has tweve vertices.

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Example Question #1 : Faces, Face Area, And Vertices

How many edges does a polyhedron with nine vertices and eleven faces have?

Possible Answers:

Correct answer:

Explanation:

By Euler's Formula, the relationship between the number of vertices , the number of faces , and the number of edges of a polyhedron is

V + F - E = 2

Set V= 9 and F = 11 and solve for :

9 +11 - E = 2

20 - E = 2

E = 18

The polyhedron has eighteen edges.

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Example Question #53 : 3 Dimensional Geometry

A rectangular prism has a height of 4 in., length of 8in., and a width of 7in. Find the volume of the prism.

Possible Answers:

$230\ \textup{in.}$

$17\ \textup{in.}$

$146\ \textup{in.}$

$224\ \textup{in.}$

Correct answer:

$224\ \textup{in.}$

Explanation:

To find the volume of a rectangular prism we use the equation of:

V=l*w*h

Now substituting in our values we get:

V=8*7*4

V=56*4

V=224

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Example Question #2 : Other 3 Dimensional Geometry

What is the volume of a triangular prism having a base of 2, a height of 8, and second height of 14?

Possible Answers:

102

112

Correct answer:

112

Explanation:

To find the volume of a triangular prism we use the equation

$V= \frac{1}{2}(b*h)*h_2$

In our case our b=2, h=8, h_2=14

Therefore,

$V=\frac{1}{2}(2*8)*14$

$V=\frac{1}{2}(16)*14$

V=8*14

V=112

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SAT II Math I : 3-Dimensional Geometry

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

All SAT II Math I Resources

Example Questions

Example Question #21 : Advanced Geometry

Example Question #51 : 3 Dimensional Geometry

Example Question #52 : 3 Dimensional Geometry

Example Question #27 : Surface Area

Example Question #22 : Surface Area

The surface area of the box is 82yd²(remember that area measurements are square units NOT cubic units)

Example Question #29 : Surface Area

Example Question #1 : Faces, Face Area, And Vertices

Example Question #1 : Faces, Face Area, And Vertices

Example Question #53 : 3 Dimensional Geometry

Example Question #2 : Other 3 Dimensional Geometry

All SAT II Math I Resources

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HIGH SCHOOL

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SAT II Math I : 3-Dimensional Geometry

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

All SAT II Math I Resources

Example Questions

Example Question #21 : Advanced Geometry

Example Question #51 : 3 Dimensional Geometry

Example Question #52 : 3 Dimensional Geometry

Example Question #27 : Surface Area

Example Question #22 : Surface Area

The surface area of the box is 82yd2 (remember that area measurements are square units NOT cubic units)

Example Question #29 : Surface Area

Example Question #1 : Faces, Face Area, And Vertices

Example Question #1 : Faces, Face Area, And Vertices

Example Question #53 : 3 Dimensional Geometry

Example Question #2 : Other 3 Dimensional 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 box is 82yd²(remember that area measurements are square units NOT cubic units)