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SAT II Math II : SAT Subject Test in Math II

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

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All SAT II Math II Resources

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

Example Questions

Example Question #12 : Geometry

Determine the area of a triangle with a base of 6, and a height of $\frac{2}{3}$ .

Possible Answers:

$\frac{1}{9}$

$\frac{2}{9}$

Correct answer:

Explanation:

Write the formula for the area of a triangle.

$A =\frac{1}{2}bh$

Substitute the base and height into the equation.

$A =\frac{1}{2}(6)(\frac{2}{3}) =2$

The answer is:

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Example Question #13 : Geometry

Find the area of a circle with a diameter of .

Possible Answers:

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

$\frac{\pi^2 x^2 }{2}$

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

$\frac{\pi x^2 }{2}$

$\frac{\pi^2 }{4}$

Correct answer:

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

Explanation:

Divide the diameter by two. This will be the radius.

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

Write the formula for the area of a circle.

$A=\pi r^2$

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

The answer is: $\frac{\pi x^2}{4}$

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Example Question #323 : Sat Subject Test In Math Ii

Hexagon

In the provided diagram, hexagon ABCDEF is regular; and are the midpoints of their respective sides. The perimeter of the hexagon is ; what is the area of Quadrilateral MBCN ?

Possible Answers:

175

$\frac{125}{4} \sqrt{3}$

$\frac{175}{4} \sqrt{3}$

$125 \sqrt{2}$

125

Correct answer:

$\frac{125}{4} \sqrt{3}$

Explanation:

Quadrilateral MBCN is a trapezoid, so we need to find the lengths of its bases and its height.

The perimeter of the hexagon is , so each side of the hexagon measures one sixth of this, or .

Construct the diameters of the hexagon, which meet at center ; construct the apothem from to $\overline{BC}$ , with point of intersection . The diagram is below:

Hexagon

The six triangles formed by the diameters are equilateral, so AO = OD = BC = 10 , and AD = AO + OD = 10+10= 20 . Quadrilateral ABCD is a trapezoid with bases of length 10 and 20. Since $\overline{MN}$ has its endpoints at the midpoints of the legs of Trapezoid ABCD , it follows that $\overline{MN}$ is a midsegment, and has as its length $MN = \frac{1}{2} (AD + BC) = \frac{1}{2} (20+ 10 ) = 15$ .

The trapezoid has bases of length and ; we now need to find its height. This is the measure of $\overline{PQ}$ , which is half the length of apothem $\overline{OP}$ . is the height of an equilateral triangle $\bigtriangleup BCO$ and, consequently, the long leg of a right triangle $\bigtriangleup BCO$ . By the 30-60-90 Theorem,

$OP = BP \cdot \sqrt{3} = \frac{1}{2} \cdot BC \cdot \sqrt{3} = \frac{1}{2} \cdot 10 \cdot \sqrt{3}=5\sqrt{3}$ .

$PQ = \frac{1}{2} \cdot OP = \frac{1}{2} \cdot 5 \sqrt{3} = \frac{5}{2} \sqrt{3}$

The area of a trapezoid of height and base lengths and is

$A = \frac{1}{2} (B+b) h$ ;

Setting $h =PQ = \frac{5}{2} \sqrt{3} , B = MN = 15, b = BC = 10$ :

$A = \frac{1}{2} (B+b) h$

$= \frac{1}{2} \cdot (15 + 10 ) \cdot \frac{5}{2} \sqrt{3}$

$= \frac{25}{2} \cdot \frac{5}{2} \sqrt{3}$

$= \frac{125}{4} \sqrt{3}$

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Example Question #11 : Area

To the nearest whole, give the area of a regular pentagon with a perimeter of fifty.

Possible Answers:

172

385

101

Correct answer:

172

Explanation:

In a regular pentagon, called Pentagon ABCDE , construct the five perpendicular segments from each vertex to its opposite side, as shown below:

Pentagon 2

The segments divide the pentagon into ten congruent triangles.

In particular, examine $\bigtriangleup OXD$ . $\overline{OD}$ , a radius of the pentagon, bisects $\angle EDC$ , which, as the interior angle of a regular pentagon, has measure $108 ^{\circ }$ ; therefore, $m \angle ODX = 54 ^{\circ }$ . $\overline{OX}$ is an apothem and therefore bisects $\overline{CD}$ ; since the pentagon has perimeter 50, $\overline{CD}$ has length one fifth of this, or 10, and XD= 5 .

Using trigonometry,

$\tan \angle ODX = \frac{OX}{XD }$ ,

or, substituting,

$\frac{OX}{5} = \tan 54 ^{\circ }$

Solving for :

$5 \cdot \frac{OX}{5} =5 \cdot \tan 54 ^{\circ }$

$OX \approx 5 \cdot (1.3764 )$

$OX \approx 6.8819$

The area of this triangle is half the product of the lengths of legs $\overline{XD}$ and $\overline{OX}$ :

$A \approx \frac{1}{2} \cdot 5 \cdot 6.8819 \approx 17.2048$

Since the pentagon comprises ten triangles of this area, multiply:

$17.2048 \cdot 10 = 172 .048$

To the nearest whole, this is 172.

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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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All SAT II Math II Resources

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #12 : Geometry

Example Question #13 : Geometry

Example Question #323 : Sat Subject Test In Math Ii

Example Question #11 : Area

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

All SAT II Math II Resources

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