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SAT II Math I : Geometry

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

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All SAT II Math I Resources

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

Example Questions

Example Question #22 : Surface Area

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

Possible Answers:

$238\pi$

$163\pi$

$300\pi$

$140\pi$

$98\pi$

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 #53 : 3 Dimensional Geometry

Figure not drawn to scale.

Box example

Find the surface area of the box above.

Possible Answers:

Surface area: 84 yd²

Surface area: 48 yd²

Surface area: 84 yd³

Surface area: 82 yd²

Surface area: 82 yd³

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 #561 : Sat Subject Test In Math I

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:

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

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

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

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

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

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 #2 : 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 #131 : 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:

$230\ \textup{in.}$

$224\ \textup{in.}$

$146\ \textup{in.}$

$17\ \textup{in.}$

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:

112

102

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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Example Question #2 : Other 3 Dimensional Geometry

If a cube has a surface area of then what is the the length of its longest diagonal?

Possible Answers:

$3\sqrt{2}$

$\frac{1}{3}\sqrt{3}$

$\sqrt{6}$

216

Correct answer:

$3\sqrt{2}$

Explanation:

The longest diagonal of a cube transverses the interior of the figure:

Cube diagonal

This distance is defined by the super pythagorean theorem :

$distance = \sqrt{a^2+b^2+c^2}$

where , , and are the length, width, and height. Because the figure is a cube, all three are the same measure, and each is the side of the cube.

We can use the given surface area to find the length of the side:

$area = 6\cdot side^2$

36 = 6 side^2

6 = side^2

$\sqrt{6} = side$

We can use this value for the side to plug into the super pythagorean theorem

$distance = \sqrt{a^2+b^2+c^2}$

$distance = \sqrt{\sqrt{6}^2+\sqrt{6}^2+\sqrt{6}^2}$

$distance = \sqrt{18}$

Which can be simplified to

$3\sqrt{2}$

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Example Question #2 : Graphing

Which of the following points is in Quadrant IV on the coordinate plane?

Possible Answers:

(4,-7)

(-4,-7)

(4,7)

Two of these points are in Quadrant IV.

(-4,7)

Correct answer:

(4,-7)

Explanation:

Quadrant IV consists of the points with positive -coordinates and negative -coordinates. Therefore (4,-7) is the correct choice.

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Example Question #1 : Parabolas And Circles

Give the axis of symmetry of the parabola of the equation

$f(x) = 3x^{2} - 9x + 16$

Possible Answers:

$x= -\frac{1}{6}$

$x = -\frac{3}{2}$

$x = \frac{3}{2}$

$x= \frac{1}{6}$

x = 16

Correct answer:

$x = \frac{3}{2}$

Explanation:

The line of symmetry of the parabola of the equation

$f(x) = ax^{2}+ bx + c$

is the vertical line

$x = -\frac{b}{2a}$

Substitute a = 3, b = -9 :

The line of symmetry is

$x = -\frac{-9}{2 \cdot 3} = \frac{9}{6} = \frac{3}{2}$

That is, the line of the equation $x = \frac{3}{2}$ .

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All SAT II Math I Resources

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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 #22 : Surface Area

Example Question #53 : 3 Dimensional Geometry

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

Example Question #561 : Sat Subject Test In Math I

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

Example Question #2 : Faces, Face Area, And Vertices

Example Question #131 : Geometry

Example Question #2 : Other 3 Dimensional Geometry

Example Question #2 : Other 3 Dimensional Geometry

Example Question #2 : Graphing

Example Question #1 : Parabolas And Circles

All SAT II Math I Resources

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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 #22 : Surface Area

Example Question #53 : 3 Dimensional Geometry

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

Example Question #561 : Sat Subject Test In Math I

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

Example Question #2 : Faces, Face Area, And Vertices

Example Question #131 : Geometry

Example Question #2 : Other 3 Dimensional Geometry

Example Question #2 : Other 3 Dimensional Geometry

Example Question #2 : Graphing

Example Question #1 : Parabolas And Circles

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)