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SAT II Math II : Geometry

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

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Example Questions

Example Question #1 : Midpoint Formula

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

Possible Answers:

(3,-2) $\displaystyle (3,-2)$

(2, -3) $\displaystyle (2, -3)$

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

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

Correct answer:

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

Explanation:

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

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

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

and $\displaystyle (x_2, y_2)=(5,7)$

Therefore, substituting these values in we get the following:

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

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

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

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Example Question #1 : Midpoint Formula

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Write the midpoint formula.

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

Substitute the points.

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

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

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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) $\displaystyle (3, 2)$ ; the other, (3, 4) $\displaystyle (3, 4)$ .

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

Possible Answers:

$69^{\circ }$ $\displaystyle 69^{\circ }$

$71^{\circ }$ $\displaystyle 71^{\circ }$

$19 ^{\circ }$ $\displaystyle 19 ^{\circ }$

$21^{\circ }$ $\displaystyle 21^{\circ }$

$34 ^{\circ }$ $\displaystyle 34 ^{\circ }$

Correct answer:

$19 ^{\circ }$ $\displaystyle 19 ^{\circ }$

Explanation:

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

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

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

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

using the known points.

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

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

The inverse tangent of this is

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

making this the angle this line forms with the $\displaystyle x$ -axis.

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

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

The inverse tangent of this is

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

making this the angle this line forms with the $\displaystyle x$ -axis.

Subtract:

Taking the inverse tangent:

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

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

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Example Question #11 : Coordinate Geometry

In the figure below, regular hexagon ABCDEF $\displaystyle ABCDEF$ has a side length of $\displaystyle 4$ . Find the y-coordinate of point $\displaystyle C$ .

Possible Answers:

3.46 $\displaystyle 3.46$

4.09 $\displaystyle 4.09$

3.99 $\displaystyle 3.99$

3.38 $\displaystyle 3.38$

Correct answer:

3.46 $\displaystyle 3.46$

Explanation:

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

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

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

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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 : Midpoint Formula

Example Question #1 : Midpoint Formula

Example Question #1 : Other Coordinate Geometry

Example Question #11 : Coordinate Geometry

All SAT II Math II Resources

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