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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 : Diameter, Radius, And Circumference

If the circumference of a circle is $2\pi +1$ , what must be the diameter?

Possible Answers:

$\frac{\pi+2}{\pi}$

$\frac{2\pi+2}{\pi}$

$\frac{2\pi+1}{\pi}$

$\frac{4\pi+4}{\pi}$

$\frac{\pi+1}{\pi}$

Correct answer:

$\frac{2\pi+1}{\pi}$

Explanation:

Write the circumference formula for the circle.

$C=2\pi r$

Substitute the circumference into the equation.

$2\pi +1 = 2\pi r$

Divide by $2\pi$ on both sides.

$\frac{2\pi +1}{ 2\pi } =\frac{ 2\pi r}{ 2\pi }$

$r=\frac{2\pi +1}{ 2\pi }$

The diameter is twice the radius.

$d=2r=2\cdot \frac{2\pi +1}{ 2\pi } = \frac{2\pi+1}{\pi}$

The answer is: $\frac{2\pi+1}{\pi}$

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Example Question #2 : Diameter, Radius, And Circumference

What is the circumference of a circle with a diameter of $\frac{\pi }{4}$ ?

Possible Answers:

$\frac{\pi^2}{2}$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\frac{\pi^2}{8}$

$\frac{\pi^2}{4}$

Correct answer:

$\frac{\pi^2}{4}$

Explanation:

Write the formula for the circumference of a circle.

$C= \pi D$

Substitute the diameter into the equation.

$C= \pi (\frac{\pi}{4}) = \frac{\pi^2}{4}$

The answer is: $\frac{\pi^2}{4}$

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Example Question #2 : Diameter, Radius, And Circumference

What is the diameter of the circle with a radius of $2\sqrt{10}$ ?

Possible Answers:

$8\sqrt{10}$

$4\sqrt{5}$

$4+\sqrt{10}$

$4\sqrt{10}$

$8\sqrt5$

Correct answer:

$4\sqrt{10}$

Explanation:

The diameter of a circle is twice the radius.

D=2r

Substitute the radius.

$D=2(2\sqrt{10}) = 4\sqrt{10}$

The answer is: $4\sqrt{10}$

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Example Question #3 : Diameter, Radius, And Circumference

Determine the radius if the circumference of a circle is $2\sqrt2$ .

Possible Answers:

$\sqrt{2\pi}$

$\frac{\sqrt2}{\pi}$

$\pi\sqrt2$

$\sqrt{2}$

$\frac{2\sqrt2}{\pi}$

Correct answer:

$\frac{\sqrt2}{\pi}$

Explanation:

Write the formula for the circumference of a circle.

$C=2\pi r$

Substitute the circumference.

$2\sqrt2=2\pi r$

Divide by $2\pi$ on both sides.

$\frac{2\sqrt2}{2\pi }=\frac{2\pi r}{2\pi }$

Reduce both sides.

The answer is: $r= \frac{\sqrt2}{\pi}$

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Example Question #3 : Diameter, Radius, And Circumference

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

Possible Answers:

$\frac{2}{5}$

$\frac{4}{9}$

$\frac{1}{6}$

$\frac{1}{3}$

$\frac{4}{3}$

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 : 2 Dimensional 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:

$BC \approx 13.2$

$BC \approx 21.8$

$BC \approx 23.2$

$BC \approx 8.1$

$BC \approx 10.6$

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:

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

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

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

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

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

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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All SAT II Math II Resources

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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 : Diameter, Radius, And Circumference

Example Question #2 : Diameter, Radius, And Circumference

Example Question #2 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #2 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #41 : 2 Dimensional Geometry

Example Question #2 : Finding Sides

Example Question #1 : Finding Sides

All SAT II Math II Resources

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