Finding Sides - SAT Subject Test in Math II

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Question

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

.

Evaluate .

Answer

By the Law of Sines,

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

Substitute 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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Question

Note: figure NOT drawn to scale.

Refer to the above diagram.

.

Which of the following expressions is equal to ?

Answer

By the Law of Sines,

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

Substitute , , and :

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

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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Question

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

Evaluate . Round to the nearest tenth, if applicable.

Answer

By the Law of Cosines,

$\left (BC \right )^{2} =\left (AB \right )^{2}+ \left (AC \right )^{2} - 2 (AB)(AC) \cos a^{\circ }$

Substitute :

$\left (BC \right )^{2} =15^{2}+10^{2} - 2 \cdot15 \cdot 10 \cos 60^{\circ }$

$\left (BC \right )^{2} =225+100 - 300 \cdot \frac{1}{2}$

$\left (BC \right )^{2} =225+100 - 150$

$\left (BC \right )^{2} = 175$

$BC = \sqrt{175} \approx 13.2$

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Question

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

Answer

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 and :

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

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

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

$x \approx 11.4$

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Question

In triangle $\Delta GHJ$ , $m\angle G = 40^{\circ }$ and $m \angle H = 80^{\circ }$ .

Which of the following statements is true about the lengths of the sides of $\Delta GHJ$ ?

Answer

In a triangle, the shortest side is opposite the angle of least measure; the longest side is opposite the angle of greatest measure. Therefore, if we order the angles, we can order their opposite sides similarly.

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

$m \angle J + m \angle G + m \angle H = 180^{\circ }$

$m \angle J + 40 ^{\circ }+ 80^{\circ } = 180^{\circ }$

$m \angle J + 120^{\circ } = 180^{\circ }$

$m \angle J =60^{\circ }$

Since

$m \angle G < m \angle J< m \angle H$ ,

we can order the lengths of their opposite sides the same way:

.

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Question

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

Answer

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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Question

Note: figure NOT drawn to scale.

Refer to the above diagram.

.

Which of the following expressions is equal to ?

Answer

By the Law of Sines,

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

Substitute , , and :

$\frac{AC}{\sin 43^{\circ }} = \frac{50}{\sin 74^{\circ } }$

Solve for :

$\frac{AC}{\sin 43^{\circ }}\cdot \sin 43^{\circ }= \frac{50}{\sin 73^{\circ } } \cdot \sin 43^{\circ }$

$AC= \frac{50 \sin 43^{\circ }}{\sin 74^{\circ } }$

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Question

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

Answer

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.

A triangle $\bigtriangleup MBC$ is formed with , , 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 , 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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Question

Regular Hexagon 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 $\overline{MN}$ .

Answer

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, .

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.

is a rectangle, so .

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

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,

Adding the segment lengths:

.

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Question

Regular Pentagon has perimeter 60.

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

Answer

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:

A triangle $\bigtriangleup BCD$ is formed with , 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 , 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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Question

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

Answer

Write the formula for the volume of the cube.

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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