SAT II Math II : Finding Sides

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

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Example Questions

Example Question #41 : 2 Dimensional Geometry

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram. 

\(\displaystyle AB = 15, AC = 10, a = 60\)

Evaluate \(\displaystyle BC\). Round to the nearest tenth, if applicable.

Possible Answers:

\(\displaystyle BC \approx 13.2\)

\(\displaystyle BC \approx 21.8\)

\(\displaystyle BC \approx 23.2\)

\(\displaystyle BC \approx 8.1\)

\(\displaystyle BC \approx 10.6\)

Correct answer:

\(\displaystyle BC \approx 13.2\)

Explanation:

By the Law of Cosines,

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

Substitute \(\displaystyle AB = 15, AC = 10, a = 60\):

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

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

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

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

\(\displaystyle BC = \sqrt{175} \approx 13.2\)

Example Question #2 : Finding Sides

In triangle \(\displaystyle \Delta GHJ\)\(\displaystyle m\angle G = 40^{\circ }\) and \(\displaystyle m \angle H = 80^{\circ }\).

Which of the following statements is true about the lengths of the sides of \(\displaystyle \Delta GHJ\) ?

Possible Answers:

\(\displaystyle HJ< GJ< GH\)

\(\displaystyle GJ < GH < HJ\)

\(\displaystyle GJ< HJ< GH\)

\(\displaystyle HJ< GH < GJ\)

\(\displaystyle GH< GJ< HJ\)

Correct answer:

\(\displaystyle HJ< GH < GJ\)

Explanation:

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 \(\displaystyle 180^{\circ }\)

\(\displaystyle m \angle J + m \angle G + m \angle H = 180^{\circ }\)

\(\displaystyle m \angle J + 40 ^{\circ }+ 80^{\circ } = 180^{\circ }\)

\(\displaystyle m \angle J + 120^{\circ } = 180^{\circ }\)

\(\displaystyle m \angle J =60^{\circ }\)

Since

\(\displaystyle m \angle G < m \angle J< m \angle H\),

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

\(\displaystyle HJ < GH < GJ\).

Example Question #1 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the above diagram.

\(\displaystyle AB = 50, a = 63, b = 43\).

Which of the following expressions is equal to \(\displaystyle AC\) ?

Possible Answers:

\(\displaystyle AC= \frac{50 \sin 43^{\circ }}{\sin 74^{\circ } }\)

\(\displaystyle AC= \frac{50 \sin 43^{\circ }}{\sin 63^{\circ } }\)

\(\displaystyle AC= \frac{50 \cos 43^{\circ }}{\cos 74^{\circ } }\)

\(\displaystyle AC= \frac{50 \cos 43^{\circ }}{\cos 63^{\circ } }\)

\(\displaystyle AC= \frac{50 \cos 74^{\circ }}{\cos 43^{\circ } }\)

Correct answer:

\(\displaystyle AC= \frac{50 \sin 43^{\circ }}{\sin 74^{\circ } }\)

Explanation:

By the Law of Sines,

\(\displaystyle \frac{AC}{\sin b^{\circ }} = \frac{AB}{\sin c^{\circ } }\).

Substitute \(\displaystyle AB = 50\)\(\displaystyle b = 43\), and \(\displaystyle c = 180 - (63+ 43) = 180 - 106 = 74\):

\(\displaystyle \frac{AC}{\sin 43^{\circ }} = \frac{50}{\sin 74^{\circ } }\)

 

Solve for \(\displaystyle AC\):

\(\displaystyle \frac{AC}{\sin 43^{\circ }}\cdot \sin 43^{\circ }= \frac{50}{\sin 73^{\circ } } \cdot \sin 43^{\circ }\)

\(\displaystyle AC= \frac{50 \sin 43^{\circ }}{\sin 74^{\circ } }\)

Example Question #41 : Geometry

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

Possible Answers:

The triangle is acute and equilateral.

The triangle is acute and scalene.

The triangle is obtuse and isosceles, but not equilateral.

The triangle is obtuse and scalene.

The triangle is acute and isosceles, but not 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 \(\displaystyle 9 \times 12 = 108\) inches, and 3 yards is equal to \(\displaystyle 3 \times 36 = 108\) inches. The triangle has sides of measure 90, 108, 108.

We compare the squares of the sides.

\(\displaystyle 90^{2}+ 108^{2} = 8,100 + 11,664 = 19,764 > 11,664 = 108^{2}\)

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.

Example Question #2 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the above diagram.

\(\displaystyle AB = 40, a = 62, b = 44\).

Which of the following expressions is equal to \(\displaystyle AC\) ?

Possible Answers:

\(\displaystyle AC= \frac{40 \sin 44^{\circ }}{\sin 62^{\circ } }\)

\(\displaystyle AC= \frac{40 \cos 74^{\circ }}{\cos 44^{\circ } }\)

\(\displaystyle AC= \frac{40 \cos 44^{\circ }}{\cos 74^{\circ } }\)

\(\displaystyle AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }\)

\(\displaystyle AC= \frac{40 \cos 44^{\circ }}{\cos 62^{\circ } }\)

Correct answer:

\(\displaystyle AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }\)

Explanation:

By the Law of Sines,

\(\displaystyle \frac{AC}{\sin b^{\circ }} = \frac{AB}{\sin c^{\circ } }\).

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

\(\displaystyle \frac{AC}{\sin 44^{\circ }} = \frac{40}{\sin 74^{\circ } }\)

We can solve for \(\displaystyle AC\):

\(\displaystyle \frac{AC}{\sin 44^{\circ }} \cdot \sin 44^{\circ }= \frac{40}{\sin 74^{\circ } } \cdot \sin 44^{\circ }\)

\(\displaystyle AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }\)

Example Question #6 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

\(\displaystyle BC = 25, a= 60, b = 45\).

Evaluate \(\displaystyle AC\).

Possible Answers:

\(\displaystyle AC= \frac{50\sqrt{6}}{ 3 }\)

\(\displaystyle AC= \frac{50\sqrt{3}}{ 3 }\)

\(\displaystyle AC= \frac{25\sqrt{6}}{ 3 }\)

\(\displaystyle AC= 25\sqrt{3}\)

\(\displaystyle AC= \frac{25\sqrt{3}}{ 3 }\)

Correct answer:

\(\displaystyle AC= \frac{25\sqrt{6}}{ 3 }\)

Explanation:

By the Law of Sines,

\(\displaystyle \frac{AC}{\sin b^{\circ }} = \frac{BC}{\sin a^{\circ } }\)

Substitute \(\displaystyle BC = 25, a= 60, b = 45\) and solve for \(\displaystyle AC\):

\(\displaystyle \frac{AC}{\sin 45^{\circ } } = \frac{25}{\sin 60^{\circ }}\)

\(\displaystyle \frac{AC}{\sin 45^{\circ } } \cdot \sin 45^{\circ }\sin 60^{\circ } = \frac{25}{\sin 60^{\circ }} \cdot \sin 45^{\circ }\sin 60^{\circ }\)

\(\displaystyle AC\sin 60^{\circ } =25 \sin 45^{\circ }\)

\(\displaystyle AC\cdot { \frac{\sqrt{3}}{2}}= 25 \cdot \frac{\sqrt{2}}{2}\)

\(\displaystyle AC\cdot { \frac{\sqrt{3}}{2}}= \frac{25 \sqrt{2}}{2}\)

\(\displaystyle AC\cdot { \frac{\sqrt{3}}{2}} \cdot \frac{2 \sqrt{3}}{3}= \frac{25 \sqrt{2}}{2} \cdot \frac{2 \sqrt{3}}{3}\)

\(\displaystyle AC \cdot \frac{6}{6}= \frac{50 \sqrt{2}\sqrt{3}}{6}\)

\(\displaystyle AC= \frac{25\sqrt{6}}{ 3 }\)

Example Question #2 : Finding Sides

Decagon

The above figure is a regular decagon. Evaluate \(\displaystyle x\) to the nearest tenth.

Possible Answers:

\(\displaystyle x \approx 11.4\)

\(\displaystyle x \approx 9.7\)

\(\displaystyle x \approx 10.4\)

\(\displaystyle x \approx 10.8\)

\(\displaystyle x \approx 11.6\)

Correct answer:

\(\displaystyle x \approx 11.4\)

Explanation:

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

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

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

\(\displaystyle c^{2}= a^{2}+ b^{2} - 2ab \cos C^{\circ }\)

with \(\displaystyle a = b = 6\) and \(\displaystyle C = 144\):

\(\displaystyle x ^{2}= 6^{2} + 6^{2} - 2 \cdot 6 \cdot 6 \cos 144^{\circ }\)

\(\displaystyle x ^{2}= 36+ 36- 72 \cos 144^{\circ }\)

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

\(\displaystyle x \approx 11.4\)

Example Question #352 : Sat Subject Test In Math Ii

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

Possible Answers:

\(\displaystyle 14.4\)

\(\displaystyle 11.3\)

\(\displaystyle 8.7\)

\(\displaystyle 9.2\)

\(\displaystyle 8.2\)

Correct answer:

\(\displaystyle 8.7\)

Explanation:

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

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

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

Pentagon 1

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

\(\displaystyle c^{2} = a^{2} + b^{2} -2 ab \cos \gamma\)

where \(\displaystyle a\) and \(\displaystyle b\) are the lengths of two sides, \(\displaystyle \gamma\) is the measure of their included angle, and \(\displaystyle c\) is the length of the third side. 

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

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

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

\(\displaystyle \approx 61.25 + 15.14\)

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

Taking the square root of both sides:

\(\displaystyle ME \approx\sqrt{ 76.39 } \approx 8.7\),

the correct choice.

Example Question #51 : Geometry

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

Possible Answers:

\(\displaystyle 79.5\)

\(\displaystyle 94.6\)

\(\displaystyle 72.4\)

\(\displaystyle 120.0\)

\(\displaystyle 90.0\)

Correct answer:

\(\displaystyle 90.0\)

Explanation:

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

Similarly, \(\displaystyle DN = NE = 30\).

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

Below is the hexagon with the midpoints \(\displaystyle M\) and \(\displaystyle N\), and with \(\displaystyle \overline{MN}\) constructed. Note that perpendiculars have been drawn to \(\displaystyle \overline{MN}\) from \(\displaystyle E\) and \(\displaystyle F\), with feet at points \(\displaystyle Y\)and \(\displaystyle X\) respectively.

Hexagon

\(\displaystyle EFXY\) is a rectangle, so \(\displaystyle XY = EF= 60\).

\(\displaystyle m \angle MBX = m \angle MBC- m \angle XBC = 120 ^{\circ } - 90 ^{\circ } = 30 ^{\circ }\)

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

\(\displaystyle MX = \frac{1}{2} \cdot MF = \frac{1}{2} \cdot 30 = 15\).

For the same reason,

\(\displaystyle YN = 1 5\)

Adding the segment lengths:

\(\displaystyle MN = MX + XY + NY = 15 + 60 + 15 = 90\).

Example Question #4 : Finding Sides

Regular Pentagon \(\displaystyle ABCDE\) has perimeter 60. 

To the nearest tenth, give the length of diagonal \(\displaystyle \overline{BD}\).

Possible Answers:

\(\displaystyle 11.6\)

\(\displaystyle 15.7\)

\(\displaystyle 19.4\)

\(\displaystyle 15 .0\)

\(\displaystyle 14.1\)

Correct answer:

\(\displaystyle 19.4\)

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 \(\displaystyle 108^{\circ }\).

The pentagon, along with diagonal \(\displaystyle \overline{BD}\), is shown below:

Pentagon 2

 

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

\(\displaystyle c^{2} = a^{2} + b^{2} -2 ab \cos \gamma\)

where \(\displaystyle a\) and \(\displaystyle b\) are the lengths of two sides, \(\displaystyle \gamma\) the measure of their included angle, and \(\displaystyle c\) the length of the side opposite that angle.

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

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

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

\(\displaystyle \approx 288 + 89.00\)

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

Taking the square root of both sides:

\(\displaystyle BD\approx\sqrt{ 377.00 } \approx 19.4\),

the correct choice.

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