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SAT II Math II : Geometry

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 : Diameter, Radius, And Circumference

Determine the radius of a circle if the circumference is $\frac{2}{3}\pi$ .

Possible Answers:

$\frac{1}{6}$

$\frac{4}{3}$

$\frac{2}{5}$

$\frac{4}{9}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

Write the formula for the circumference of a circle.

$C=2\pi r$

Substitute the circumference.

$\frac{2}{3}\pi=2\pi r$

Multiply by $\frac{1}{2\pi}$ on both sides to isolate .

$\frac{2}{3}\pi\cdot \frac{1}{2\pi}=2\pi r \cdot \frac{1}{2\pi}$

$r= \frac{1}{3}$

The radius is: $\frac{1}{3}$

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Example Question #2 : Diameter, Radius, And Circumference

Determine the diameter if the radius of a circle is $\frac{7}{3}$ .

Possible Answers:

$\frac{49}{9}$

$\frac{14}{3}$

$\frac{7}{6}$

$\frac{2}{7}$

$\frac{7}{49}$

Correct answer:

$\frac{14}{3}$

Explanation:

The diameter is double the radius. Multiply the radius by two.

D= 2r

$D= 2(\frac{7}{3}) = \frac{14}{3}$

The answer is: $\frac{14}{3}$

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Example Question #3 : Diameter, Radius, And Circumference

Determine the circumference of a circle with a radius of $\frac{2}{\pi}$ .

Possible Answers:

$2\pi^2$

$4\pi^2$

$2\pi$

Correct answer:

Explanation:

The circumference of a circle is: $C=2\pi r$

Substitute the radius.

$C=2\pi (\frac{2}{\pi})$

The answer is:

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Example Question #41 : Geometry

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

AB = 15, AC = 10, a = 60

Evaluate . Round to the nearest tenth, if applicable.

Possible Answers:

$BC \approx 8.1$

$BC \approx 13.2$

$BC \approx 10.6$

$BC \approx 21.8$

$BC \approx 23.2$

Correct answer:

$BC \approx 13.2$

Explanation:

By the Law of Cosines,

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

Substitute AB = 15, AC = 10, a = 60 :

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

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

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

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

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

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Example Question #2 : Finding Sides

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 $\Delta GHJ$ ?

Possible Answers:

HJ< GJ<GH

GJ < GH <HJ

GJ<HJ<GH

HJ< GH < GJ

GH<GJ< HJ

Correct answer:

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 $180^{\circ }$ ,

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

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

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

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

Since

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

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

HJ < GH < GJ .

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Example Question #1 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the above diagram.

AB = 50, a = 63, b = 43 .

Which of the following expressions is equal to ?

Possible Answers:

$AC= \frac{50 \cos 43^{\circ }}{\cos 63^{\circ } }$

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

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

$AC= \frac{50 \sin 43^{\circ }}{\sin 63^{\circ } }$

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

Correct answer:

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

Explanation:

By the Law of Sines,

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

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

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

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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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 acute and scalene.

The triangle is obtuse 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 #3 : 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:

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

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

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

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

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

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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All SAT II Math II Resources

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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 #1 : Diameter, Radius, And Circumference

Example Question #2 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #41 : Geometry

Example Question #2 : Finding Sides

Example Question #1 : Finding Sides

Example Question #1 : Finding Sides

Example Question #3 : Finding Sides

Example Question #6 : Finding Sides

Example Question #2 : Finding Sides

All SAT II Math II Resources

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