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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 #131 : Geometry

What is the distance between the point (1,2) and (8,5)?

Possible Answers:

13.9

$\sqrt58$

Correct answer:

$\sqrt58$

Explanation:

For this question we will use the distance formula to solve.

$d=\sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

In our case (x_1, y_1)=(1,2)

and (x_2, y_2)= (8,5)

Substituting these values in we get the following

$d=\sqrt((8-1)^2+(5-2)^2)$

$d=\sqrt(7^2+3^2)$

$d=\sqrt(49+9)$

$d=\sqrt(58)$

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Example Question #1 : Distance Formula

The points A=(-2,0), B=(0,3), and C=(0,0) makes a triangle. What is the distance between point A and point B?

Possible Answers:

$\sqrt13$

$\sqrt15$

Correct answer:

$\sqrt13$

Explanation:

For this question we need to use the distance formula for points A and B.

Point A will be our (x_1, y_1) and point B will be (x_2, y_2)

Now we substitute these values into the following:

$d=\sqrt((0--2)^2+(3-0)^2)$

$d=\sqrt(2^2+3^2)$

$d=\sqrt(4+9)$

$d=\sqrt13$

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Example Question #441 : Sat Subject Test In Math Ii

Find the distance between the two points (2,7) and (4,6).

Possible Answers:

$\sqrt{3}$

$\sqrt{5}$

$\sqrt{205}$

$\sqrt{17}$

Correct answer:

$\sqrt{5}$

Explanation:

The distance between two points is found using the formula $d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

For this problem the values are as follows:

$x_{1}=2$

$x_{2}=4$

$y_{1}=7$

$y_{2}=6$

Input the values into the formula and simplify

$d=\sqrt{(4-2)^2+(7-6)^2}=\sqrt{(2)^2+(1)^2}=\sqrt{4+1}=\sqrt{5}$

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Example Question #1 : Midpoint Formula

Find the point halfway between points A and B.

A=(19,54)

B=(67,34)

Possible Answers:

(86,88)

(42,45)

(43,44)

(44,43)

Correct answer:

(43,44)

Explanation:

Find the point halfway between points A and B.

A=(19,54)

B=(67,34)

We are going to need to use midpoint formula. If you ever have difficulty recalling midpoint formula, try to recall that it is basically taking two averages. One average is the average of your x values, the other average is the average of your y values.

$Mid=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Now we plug and chug!

$Mid=(\frac{19+67}{2},\frac{54+34}{2})=(\frac{86}{2},\frac{88}{2})=(43,44)$

So our answer is (43,44)

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Example Question #2 : Midpoint Formula

What is the coordinates of the point exactly half way between (-2, -3) and (5, 7)?

Possible Answers:

$\left(2, \frac{-3}{2}\right)$

(2, -3)

(3,-2)

$\left(\frac{3}{2}, 2\right)$

Correct answer:

$\left(\frac{3}{2}, 2\right)$

Explanation:

We need to use the midpoint formula to solve this question.

$midpoint=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

In our case (x_1, y_1)=(-2,-3)

and (x_2, y_2)=(5,7)

Therefore, substituting these values in we get the following:

$midpoint=(\frac{-2+5}{2}, \frac{-3+7}{2})$

$midpoint= (\frac{3}{2}, \frac{4}{2})$

$(\frac{3}{2}, 2)$

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Example Question #3 : Midpoint Formula

Find the midpoint between (2,3) and (-1,9) .

Possible Answers:

$(\frac{3}{2}, -3)$

$(\frac{1}{2}, -6)$

$(\frac{1}{2}, 6)$

$(\frac{3}{2}, 3)$

$(\frac{1}{2}, -3)$

Correct answer:

$(\frac{1}{2}, 6)$

Explanation:

Write the midpoint formula.

$M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substitute the points.

$M = (\frac{2+(-1)}{2}, \frac{3+9}{2}) = (\frac{1}{2}, 6)$

The answer is: $(\frac{1}{2}, 6)$

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Example Question #1 : Other Coordinate Geometry

On the coordinate plane, two lines intersect at the origin. One line passes through the point (3, 2) ; the other, (3, 4) .

Give the measures of the acute angles they form at their intersection (nearest degree).

Possible Answers:

$34 ^{\circ }$

$71^{\circ }$

$21^{\circ }$

$19 ^{\circ }$

$69^{\circ }$

Correct answer:

$19 ^{\circ }$

Explanation:

If $\theta$ is the measure of the angles that two lines with slopes $m_{1}$ and $m_{2}$ form, then

$\tan \theta = \frac{m_{1}-m _{2}}{1+m_{1}m_{2}}$ ,

The slopes of the lines can be found by applying the slope formula

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

using the known points.

For the first line, set $x_{1}= y_{1} = 0, x_{2}= 3, y_{2} = 4$ :

$m_{1} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{4-0}{3-0} = \frac{4}{3}$

The inverse tangent of this is

$\tan^{-1} \frac{4}{3} \approx 53.1^{\circ }$ ,

making this the angle this line forms with the -axis.

For the second line, set $x_{1}= y_{1} = 0, x_{2}= 3, y_{2} = 2$ :

$m_{2} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{2-0}{3-0} = \frac{2}{3}$

The inverse tangent of this is

$\tan^{-1} \frac{2}{3} \approx 33.7^{\circ }$

making this the angle this line forms with the -axis.

Subtract:

Taking the inverse tangent:

$\theta \approx 53.1^{\circ } - 33.7^{\circ } \approx 19.4 ^{\circ }$ .

Rounding to the nearest degree, this is $19 ^{\circ }$ .

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Example Question #2 : Other Coordinate Geometry

In the figure below, regular hexagon ABCDEF has a side length of . Find the y-coordinate of point .

Possible Answers:

3.46

3.99

3.38

4.09

Correct answer:

3.46

Explanation:

From the given information, we know that the coordinate for must be (0, 8) .

Recall that the interior angle of a regular hexagon is $120^{\circ}$ . Thus, we can draw in the following 30-60-90 triangle.

Since we know that this is a 30-60-90 triangle, we know that the sides must be as marked, in the ratio of $1:\sqrt3:2$ . Thus, the y-coordinate of must be 3.46 .

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Example Question #1 : Finding Sides With Trigonometry

The area of a regular pentagon is 1,000. Give its perimeter to the nearest whole number.

Possible Answers:

121

118

107

199

Correct answer:

121

Explanation:

A regular pentagon can be divided into ten congruent triangles by its five radii and its five apothems. Each triangle has the following shape:

Thingy_1

The area of one such triangle is $\frac{1}{2}ab$ , so the area of the entire pentagon is ten times this, or $10 \cdot \frac{1}{2}ab = 5ab$ .

The area of the pentagon is 1,000, so

5ab = 1,000

ab = 200

Also,

$\frac{a}{b} = \tan 54^{\circ }$ , or equivalently, $a=b \tan 54^{\circ }$ , so we solve for in the equation:

$b \tan 54^{ \circ } \cdot b = 200$

$b^{2} \tan 54^{ \circ } = 200$

$b^{2} = \frac{200}{\tan 54^{ \circ }} \approx \frac{200}{ 1.3764} \approx 145.3$

$b \approx \sqrt{145.3} \approx 12.1$

The perimeter is ten times this, or 121.

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Example Question #441 : Sat Subject Test In Math Ii

The area of a regular dodecagon (twelve-sided polygon) is 600. Give its perimeter to the nearest whole number.

Possible Answers:

102

129

147

Correct answer:

Explanation:

A regular dodecagon can be divided into twenty-four congruent triangles by its twelve radii and its twelve apothems, each of which is shaped as shown:

Thingy_2

The area of one such triangle is $\frac{1}{2}ab$ , so the area of the entire dodecagon is twenty-four times this, or

$24 \cdot \frac{1}{2}ab = 12ab$ .

The area of the dodecagon is 600, so

12ab = 600 , or

ab = 50 .

Also,

$\frac{a}{b} = \tan 75^{\circ }$ , or equivalently, $a=b \tan 75^{\circ }$ , so solve for in the equation

$b \tan 75^{ \circ } \cdot b =50$

Solve for :

$b^{2} \cdot \tan 75^{ \circ } = 50$

$b^{2} = \frac{50}{ \tan 75^{ \circ }} \approx \frac{50}{ 3.7321} \approx 13.4$

$b \approx \sqrt{13.4} \approx 3.66$

The perimeter is twenty-four times this:

$P \approx 3.66 \cdot 24 \approx 88$

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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 #131 : Geometry

Example Question #1 : Distance Formula

Example Question #441 : Sat Subject Test In Math Ii

Example Question #1 : Midpoint Formula

Example Question #2 : Midpoint Formula

Example Question #3 : Midpoint Formula

Example Question #1 : Other Coordinate Geometry

Example Question #2 : Other Coordinate Geometry

Example Question #1 : Finding Sides With Trigonometry

Example Question #441 : Sat Subject Test In Math Ii

All SAT II Math II Resources

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