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SAT II Math II : 2-Dimensional Geometry

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

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2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : 2 Dimensional Geometry

Find the area of a square with a length of 2x^3 .

Possible Answers:

2x^6

4x^3

2x^9

4x^6

4x^9

Correct answer:

4x^6

Explanation:

The formula for the area of a square is:

A= s^2

Substitute the side length.

A= (2x^3)^2 = (2x^3)(2x^3) = 4x^6

The answer is: 4x^6

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Example Question #22 : Geometry

Find the area of a triangle with a base of x^2 and a height of .

Possible Answers:

2x^3

4x^3

4x^2

$\frac{x^2+4x}{2}$

8x^3

Correct answer:

2x^3

Explanation:

Write the formula for the area of a triangle.

$A= \frac{bh}{2}$

Substitute the base and height.

$A= \frac{x^2(4x)}{2} = \frac{4x^3}{2} = 2x^3$

The answer is: 2x^3

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Example Question #22 : 2 Dimensional Geometry

Determine the area of a circle with a radius of $\frac{3}{5}\pi$ .

Possible Answers:

$\frac{9}{25}\pi ^2$

$\frac{9}{5}\pi ^2$

$\frac{9}{5}\pi ^3$

$\frac{9}{25}\pi ^3$

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

Correct answer:

$\frac{9}{25}\pi ^3$

Explanation:

Write the formula of the area of a circle.

$A= \pi r^2$

Substitute the radius.

$A= \pi (\frac{3}{5}\pi)^2=\pi (\frac{3}{5}\pi)(\frac{3}{5}\pi)$

The answer is: $\frac{9}{25}\pi ^3$

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Example Question #24 : Geometry

Determine the area of a circle with a diameter of .

Possible Answers:

$9x\pi$

$18x^2\pi$

$36x^2\pi$

$9x^2 \pi$

$6x^2\pi$

Correct answer:

$9x^2 \pi$

Explanation:

Write the formula for the area of a circle.

$A= \pi r^2$

The radius is half the diameter, .

Substitute the radius into the equation.

$A= \pi (3x)^2 = 9x^2 \pi$

The answer is: $9x^2 \pi$

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Example Question #1 : Perimeter

Thingy

Refer to the above figure. Quadrilateral BCDE is a square. What is the perimeter of Polygon ABCDE ?

Possible Answers:

Correct answer:

Explanation:

$\overline{BE }$ is both one side of Square BCDE and the hypotenuse of $\Delta ABE$ ; its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

$BE = \sqrt{5^{2}+12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$ .

Since Square BCDE has four congruent sides, each side has length 13.

The perimeter of Polygon ABCDE is

P = AB + BC + BC + CD + DE = 5 + 13 + 13 + 13 + 12 = 56

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Example Question #2 : Perimeter

Garden

Note: Figure NOT drawn to scale

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path (in orange) seven feet wide throughout. What is the perimeter of the garden?

Possible Answers:

$146 \textrm{ ft}$

$306 \textrm{ ft}$

$132 \textrm{ ft}$

$292\textrm{ ft}$

$264\textrm{ ft}$

Correct answer:

$264\textrm{ ft}$

Explanation:

The inner rectangle, which represents the garden, has length and width $(100 - 2\times 7) = 86$ feet and $(60 - 2\times 7) = 46$ feet, respectively, so its perimeter is

P = 86+ 86 + 46 + 46 = 264 feet.

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Example Question #3 : Perimeter

Garden

Refer to the above figure, which shows a square garden (in green) surrounded by a dirt path (in orange). The dirt path is seven feet wide throughout. Which of the following polynomials gives the perimeter of the garden in feet?

Possible Answers:

14x-49

28x - 196

4x - 28

4x - 56

$x^{2} - 49$

Correct answer:

4x - 56

Explanation:

The sidelength of the garden, in feet, is $2 \times 7 = 14$ feet less than that of the entire lot, or

S = x-14 ;

The perimeter, in feet, of the garden is four times this:

$P = 4S = 4 \left (x-14 \right ) = 4x - 56$

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Example Question #22 : Geometry

You have a pentagonal-shaped lot with side lengths of 120ft ,30ft, 55ft, 60ft, and a longest side which is triple the length of the third longest side. What is the perimeter around the lot?

Possible Answers:

180ft

445ft

265ft

655ft

Correct answer:

445ft

Explanation:

You have a pentagonal-shaped lot with side lengths of 120ft,30ft, 55ft, 60 ft, and a longest side which is triple the length of the third longest side. What is the perimeter around the lot?

So, we have a five-sided lot and are given 4 sides, and the means to find the 5th.

The longest side is triple the length of the third longest side:

The third longest side is 60 feet, therefore, the longest side is 180 ft

Find the perimeter by adding up all the sides:

120ft+30ft+55ft+60ft+180ft=445ft

So our answer is 445 feet

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Example Question #2 : Perimeter

Find the circumference of a circle with diameter of 15.

Possible Answers:

$56.25\pi$

$30\pi$

$49\pi$

$15\pi$

$35\pi$

Correct answer:

$15\pi$

Explanation:

The circumference of a circle is denoted by the formula $C=2*\pi*r$ where r is the radius of the circle. The problem gives us the diameter of the circle which must be used to find the radius.

$radius=\frac{Diameter}{2}=\frac{15}{2}=7.5$

Plug the radius into the formula for the area of a circle

$C=2*\pi*7.5=15\pi$

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Example Question #23 : Geometry

If the sides of a triangle are -2x+3 , , and -2(3x-2) , what is the perimeter?

Possible Answers:

2x-1

$\frac{13}{2}x+7$

x+7

13x+7

$\frac{1}{2}x+\frac{7}{2}$

Correct answer:

x+7

Explanation:

The perimeter of a triangle is the sum of all three sides of the triangle.

Add all the side lengths.

(-2x+3) +( 9x)+ [-2(3x-2)]

Expand the terms in the bracket and remove all parentheses.

-2x+3+9x-6x+4

Combine like-terms.

The answer is: x+7

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SAT II Math II : 2-Dimensional Geometry

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

All SAT II Math II Resources

Example Questions

Example Question #21 : 2 Dimensional Geometry

Example Question #22 : Geometry

Example Question #22 : 2 Dimensional Geometry

Example Question #24 : Geometry

Example Question #1 : Perimeter

Example Question #2 : Perimeter

Example Question #3 : Perimeter

Example Question #22 : Geometry

Example Question #2 : Perimeter

Example Question #23 : Geometry

All SAT II Math II Resources

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