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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 #1 : Finding Sides

Which of the following describes a triangle with sides of length 9 feet, 3 yards, and 90 inches?

Possible Answers:

The triangle is obtuse and isosceles, but not equilateral.

The triangle is obtuse and scalene.

The triangle is acute and scalene.

The triangle is acute and isosceles, but not equilateral.

The triangle is acute and equilateral.

Correct answer:

The triangle is acute and isosceles, but not equilateral.

Explanation:

One yard is equal to three feet; One foot is equal to twelve inches. Therefore, 9 feet is equal to $9 \times 12 = 108$ inches, and 3 yards is equal to $3 \times 36 = 108$ inches. The triangle has sides of measure 90, 108, 108.

We compare the squares of the sides.

$90^{2}+ 108^{2} = 8,100 + 11,664 = 19,764 > 11,664 = 108^{2}$

The sum of the squares of the two smaller sidelengths exceeds that of the third, so the triangle is acute.

The correct response is acute and isosceles.

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Example Question #2 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the above diagram.

AB = 40, a = 62, b = 44 .

Which of the following expressions is equal to ?

Possible Answers:

$AC= \frac{40 \sin 44^{\circ }}{\sin 62^{\circ } }$

$AC= \frac{40 \cos 74^{\circ }}{\cos 44^{\circ } }$

$AC= \frac{40 \cos 44^{\circ }}{\cos 74^{\circ } }$

$AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }$

$AC= \frac{40 \cos 44^{\circ }}{\cos 62^{\circ } }$

Correct answer:

$AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }$

Explanation:

By the Law of Sines,

$\frac{AC}{\sin b^{\circ }} = \frac{AB}{\sin c^{\circ } }$ .

Substitute AB = 40 , b = 44 , and c = 180 - (62+ 44) = 180 - 106 = 74 :

$\frac{AC}{\sin 44^{\circ }} = \frac{40}{\sin 74^{\circ } }$

We can solve for :

$\frac{AC}{\sin 44^{\circ }} \cdot \sin 44^{\circ }= \frac{40}{\sin 74^{\circ } } \cdot \sin 44^{\circ }$

$AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }$

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Example Question #6 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

BC = 25, a= 60, b = 45 .

Evaluate .

Possible Answers:

$AC= \frac{50\sqrt{6}}{ 3 }$

$AC= \frac{50\sqrt{3}}{ 3 }$

$AC= \frac{25\sqrt{6}}{ 3 }$

$AC= 25\sqrt{3}$

$AC= \frac{25\sqrt{3}}{ 3 }$

Correct answer:

$AC= \frac{25\sqrt{6}}{ 3 }$

Explanation:

By the Law of Sines,

$\frac{AC}{\sin b^{\circ }} = \frac{BC}{\sin a^{\circ } }$

Substitute BC = 25, a= 60, b = 45 and solve for :

$\frac{AC}{\sin 45^{\circ } } = \frac{25}{\sin 60^{\circ }}$

$\frac{AC}{\sin 45^{\circ } } \cdot \sin 45^{\circ }\sin 60^{\circ } = \frac{25}{\sin 60^{\circ }} \cdot \sin 45^{\circ }\sin 60^{\circ }$

$AC\sin 60^{\circ } =25 \sin 45^{\circ }$

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

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

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

$AC \cdot \frac{6}{6}= \frac{50 \sqrt{2}\sqrt{3}}{6}$

$AC= \frac{25\sqrt{6}}{ 3 }$

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Example Question #2 : Finding Sides

Decagon

The above figure is a regular decagon. Evaluate to the nearest tenth.

Possible Answers:

$x \approx 11.4$

$x \approx 9.7$

$x \approx 10.4$

$x \approx 10.8$

$x \approx 11.6$

Correct answer:

$x \approx 11.4$

Explanation:

Two sides of the triangle formed measure 6 each; the included angle is one angle of the regular decagon, which measures

$m = \left [\frac{180(10-2)}{10} \right ] ^{\circ }= 144^{\circ }$ .

Since we know two sides and the included angle of the triangle in the diagram, we can apply the Law of Cosines,

$c^{2}= a^{2}+ b^{2} - 2ab \cos C^{\circ }$

with a = b = 6 and C = 144 :

$x ^{2}= 6^{2} + 6^{2} - 2 \cdot 6 \cdot 6 \cos 144^{\circ }$

$x ^{2}= 36+ 36- 72 \cos 144^{\circ }$

$x ^{2} \approx 72 - 72 \left ( -0.8090\right )$

$x \approx 11.4$

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Example Question #352 : Sat Subject Test In Math Ii

Regular Pentagon ABCDE has perimeter 35. $\overline{AB }$ has as its midpoint; segment $\overline{ME}$ is drawn. To the nearest tenth, give the length of $\overline{ME}$ .

Possible Answers:

14.4

11.3

8.7

9.2

8.2

Correct answer:

8.7

Explanation:

The perimeter of the regular pentagon is 35, so each side measures one fifth of this, or 7. Also, since is the midpoint of $\overline{AB}$ , $AM = BM = \frac{1}{2}AB = \frac{1}{2} \cdot 7= 3.5$ .

Also, each interior angle of a regular pentagon measures $108^{\circ }$ .

Below is the pentagon in question, with indicated and $\overline{ME}$ constructed; all relevant measures are marked.

Pentagon 1

A triangle $\bigtriangleup MBC$ is formed with AM = 3.5 , AE= 7 , and included angle measure $\angle A= 108^{\circ }$ . The length of the remaining side can be calculated using the law of cosines:

$c^{2} = a^{2} + b^{2} -2 ab \cos \gamma$

where and are the lengths of two sides, $\gamma$ is the measure of their included angle, and is the length of the third side.

Setting $a = AM = 3.5, b=AE= 7, \gamma= m \angle A = 108^{\circ }$ , and c = ME , substitute and evaluate :

$(ME)^{2} = 3.5 ^{2} + 7^{2} -2 (3.5) (7) \cos 108^{\circ }$

$\approx 12.25 + 49 -49(-0.3090)$

$\approx 61.25 + 15.14$

$(ME)^{2} \approx 76.39$ ;

Taking the square root of both sides:

$ME \approx\sqrt{ 76.39 } \approx 8.7$ ,

the correct choice.

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Example Question #51 : Geometry

Regular Hexagon ABCDEF has perimeter 360. $\overline{AF}$ and $\overline{DE}$ have and as midpoints, respectively; segment $\overline{MN}$ is drawn. To the nearest tenth, give the length of $\overline{MN}$ .

Possible Answers:

79.5

94.6

72.4

120.0

90.0

Correct answer:

90.0

Explanation:

The perimeter of the regular hexagon is 360, so each side measures one sixth of this, or 60. Since is the midpoint of $\overline{AB}$ , $AM = MF = \frac{1}{2}AF= \frac{1}{2} \cdot 60 = 30$ .

Similarly, DN = NE = 30 .

Also, each interior angle of a regular hexagon measures $120 ^{\circ }$ .

Below is the hexagon with the midpoints and , and with $\overline{MN}$ constructed. Note that perpendiculars have been drawn to $\overline{MN}$ from and , with feet at points and respectively.

Hexagon

EFXY is a rectangle, so XY = EF= 60 .

$m \angle MBX = m \angle MBC- m \angle XBC = 120 ^{\circ } - 90 ^{\circ } = 30 ^{\circ }$ .

This makes $\overline{MX}$ and $\overline{MF}$ the short leg and hypotenuse of a 30-60-90 triangle; as a consequence,

$MX = \frac{1}{2} \cdot MF = \frac{1}{2} \cdot 30 = 15$ .

For the same reason,

YN = 1 5

Adding the segment lengths:

MN = MX + XY + NY = 15 + 60 + 15 = 90 .

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Example Question #4 : Finding Sides

Regular Pentagon ABCDE has perimeter 60.

To the nearest tenth, give the length of diagonal $\overline{BD}$ .

Possible Answers:

11.6

15.7

19.4

15 .0

14.1

Correct answer:

19.4

Explanation:

The perimeter of the regular pentagon is 60, so each side measures one fifth of this, or 12. Also, each interior angle of a regular pentagon measures $108^{\circ }$ .

The pentagon, along with diagonal $\overline{BD}$ , is shown below:

Pentagon 2

A triangle $\bigtriangleup BCD$ is formed with AB=BC = 16 , and included angle measure $\angle B= 108^{\circ }$ . The length of the remaining side can be calculated using the Law of Cosines:

$c^{2} = a^{2} + b^{2} -2 ab \cos \gamma$

where and are the lengths of two sides, $\gamma$ the measure of their included angle, and the length of the side opposite that angle.

Setting $a = BC= 12 , b= CD = 12, \gamma = m \angle C = 108^{\circ }$ , and c= BD , substitute and evaluate :

$(BD)^{2} = 12 ^{2} + 12 ^{2} -2 (12) (12) \cos 108^{\circ }$

$\approx 144+144 -288 (-0.3090)$

$\approx 288 + 89.00$

$(BD)^{2} \approx 377.00$

Taking the square root of both sides:

$BD\approx\sqrt{ 377.00 } \approx 19.4$ ,

the correct choice.

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Example Question #51 : 2 Dimensional Geometry

Given a cube, if the volume is 100 feet cubed, what must be the side?

Possible Answers:

$10\sqrt[3]{3}$

$\sqrt[3]{100}$

$\sqrt{\frac{100}{3}}$

$10\sqrt3$

Correct answer:

$\sqrt[3]{100}$

Explanation:

Write the formula for the volume of the cube.

V=s^3

To solve for , cube root both sides.

$\sqrt[3]{V}=\sqrt[3]{s^3}$

$s=\sqrt[3]{V}$

Substitute the volume.

The answer is: $\sqrt[3]{100}$

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Example Question #1 : Angles

Solve for and .

(Figure not drawn to scale).

Possible Answers:

$\small x=15^o;\ y=52.5^o$

$\small x=15^o;\ y=37.5^o$

$\small x=7.5^o;\ y=18.75^o$

$\small x=15^o;\ y=7.5^o$

Correct answer:

$\small x=15^o;\ y=37.5^o$

Explanation:

The angles containing the variable all reside along one line, therefore, their sum must be 180^o .

$\small 5x+4x+3x=180^o$

$\small 12x=180^o$

$\small x=15^o$

Because and are opposite angles, they must be equal.

$\small 2y=5x$

$\small x=15^o$

$\small 2y=5(15^o)=75^o$

$\small y=\frac{75^o}{2}=37.5^o$

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Example Question #2 : Finding Angles

What angle do the minute and hour hands of a clock form at 6:15?

Possible Answers:

$95^{\circ }$

$97 \frac{1}{2}^{\circ }$

$90^{\circ }$

$87 \frac{1}{2}^{\circ }$

$92 \frac{1}{2}^{\circ }$

Correct answer:

$97 \frac{1}{2}^{\circ }$

Explanation:

There are twelve numbers on a clock; from one to the next, a hand rotates $\left (\frac{360}{12} \right )^{\circ } = 30^{\circ }$ . At 6:15, the minute hand is exactly on the "3" - that is, on the $\left (30 \times 3 \right )^{\circ }= 90^{\circ }$ position. The hour hand is one-fourth of the way from the "6" to the "7" - that is, on the $\left (6 \frac{1}{4} \times 30 \right )^{\circ } = 187 \frac{1}{2} ^{\circ }$ position. Therefore, the difference is the angle they make:

$187 \frac{1}{2} ^{\circ } - 90^{\circ } = 97 \frac{1}{2}^{\circ }$ .

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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 #1 : Finding Sides

Example Question #2 : Finding Sides

Example Question #6 : Finding Sides

Example Question #2 : Finding Sides

Example Question #352 : Sat Subject Test In Math Ii

Example Question #51 : Geometry

Example Question #4 : Finding Sides

Example Question #51 : 2 Dimensional Geometry

Example Question #1 : Angles

Example Question #2 : Finding Angles

All SAT II Math II Resources

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