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SAT II Math II : Geometry

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Example Questions

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Example Question #8 :Finding Sides

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

Possible Answers:

14.4

11.3

8.2

9.2

8.7

Correct answer:

8.7

Explanation:

The perimeter of the regular pentagon is 35, so each side measures one fifth of this, or 7. Also, sinceis 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, withindicated 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:

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

whereandare the lengths of two sides, $\gamma$ is the measure of their included angle, andis 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 #9 :Finding Sides

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

Possible Answers:

94.6

120.0

72.4

90.0

79.5

Correct answer:

90.0

Explanation:

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

同样的, DN = NE = 30 .

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

Below is the hexagon with the midpointsand, and with $\overline{MN}$ constructed. Note that perpendiculars have been drawn to $\overline{MN}$ fromand, with feet at pointsandrespectively.

Hexagon

EFXY 是一个矩形,所以呢 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 #10 :Finding Sides

Regular Pentagon ABCDE has perimeter 60.

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

Possible Answers:

15.7

19.4

15 .0

11.6

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:

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

whereandare the lengths of two sides, $\gamma$ the measure of their included angle, andthe 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 :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 :Finding Angles

Solve forand.

(Figure not drawn to scale).

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

$\small 12x=180^o$

$\small x=15^o$

Becauseandare 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:

$90^{\circ }$

$95^{\circ }$

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

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

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

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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Example Question #3 :Finding Angles

In triangle $\Delta ABC$ , $m\angle A = 50^{\circ }$ and $m \angle B = 80^{\circ }$ . Which of the following describes the triangle?

Possible Answers:

$\Delta ABC$ is acute and isosceles.

$\Delta ABC$ is obtuse and scalene.

$\Delta ABC$ is acute and scalene.

$\Delta ABC$ is obtuse and isosceles.

None of the other responses is correct.

Correct answer:

$\Delta ABC$ is acute and isosceles.

Explanation:

Since the measures of the three interior angles of a triangle must total $180^{\circ }$ ,

$m \angle C + m \angle A + m \angle B = 180^{\circ }$

$m \angle C + 50^{\circ } + 80^{\circ } = 180^{\circ }$

$m \angle C + 130^{\circ } = 180^{\circ }$

$m \angle C = 50^{\circ }$

All three angles have measure less than $90^{\circ }$ , making the triangle acute. Also, by the Isosceles Triangle Theorem, since $m \angle A= m \angle C = 50^{\circ }$ , BC = AB ; the triangle has two congruent sides and is isosceles.

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Example Question #4 :Finding Angles

In $\Delta DEF$ , $\angle D$ and $\angle E$ are complementary, and $m\angle D > m \angle E$ . Which of the following is true of $\Delta DEF$ ?

Possible Answers:

$\Delta DEF$ is acute and isosceles.

$\Delta DEF$ is acute and scalene.

$\Delta DEF$ is right and isosceles.

$\Delta DEF$ is right and scalene.

None of the other responses is correct.

Correct answer:

$\Delta DEF$ is right and scalene.

Explanation:

$\angle D$ and $\angle E$ are complementary, so, by definition, $m\angle D + m\angle E = 90^{\circ }$ .

Since the measures of the three interior angles of a triangle must total $180^{\circ }$ ,

$m \angle F + m \angle D + m \angle E = 180^{\circ }$

$m \angle F + 90^{\circ } = 180^{\circ }$

$m \angle F = 90^{\circ }$

$\angle F$ is a right angle, so $\Delta DEF$ is a right triangle.

$\angle D$ and $\angle E$ must be acute, so neither is congruent to $\angle F$ ; also, $\angle D$ and $\angle E$ are not congruent to each other. Therefore, all three angles have different measure. Consequently, all three sides have different measure, and $\Delta DEF$ is scalene.

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Example Question #5 :Finding Angles

Decagon

The above figure is a regular decagon. Evaluate $m \angle ACB$ .

Possible Answers:

$15^{\circ }$

$12^{\circ }$

$24^{\circ }$

$20^{\circ }$

$18^{\circ }$

Correct answer:

$18^{\circ }$

Explanation:

As an interior angle of a regular decagon, $\angle B$ measures

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

Since $\overline{AB}$ and $\overline{BC}$ are two sides of a regular polygon, they are congruent. Therefore, by the Isosceles Triangle Theorem,

$m \angle ACB = m \angle CAB$

The sum of the measures of a triangle is $180^{\circ }$ , so

$m \angle ACB + m \angle CAB + m \angle ABC =180^{\circ }$

$m \angle ACB + m \angle ACB + 144 ^{\circ } =180^{\circ }$

$2 m \angle ACB + 144 ^{\circ } =180^{\circ }$

$2 m \angle ACB =36^{\circ }$

$m \angle ACB =18^{\circ }$

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Example Question #6 :Finding Angles

Hexagon

The above hexagon is regular. What is?

Possible Answers:

x = 64

x = 69

None of the other responses is correct.

x = 74

x = 79

Correct answer:

x = 79

Explanation:

Two of the angles of the quadrilateral formed are angles of a regular hexagon, so each measures

$\frac{180^{\circ } (6-2)}{6}= 120^{\circ }$ .

The four angles of the quadrilateral are $120 ^{\circ }, 120^{\circ }, 41^{\circ }, x^{\circ }$ . Their sum is $360^{\circ }$ , so we can set up, and solve for在方程:

x + 120 + 120 + 41 =360

x + 281 =360

x = 79

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SAT II Math II : Geometry

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

All SAT II Math II Resources

Example Questions

Example Question #8 :Finding Sides

Example Question #9 :Finding Sides

Example Question #10 :Finding Sides

Example Question #51 :Geometry

Example Question #1 :Finding Angles

Example Question #2 :Finding Angles

Example Question #3 :Finding Angles

Example Question #4 :Finding Angles

Example Question #5 :Finding Angles

Example Question #6 :Finding Angles

All SAT II Math II Resources

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