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:

\displaystyle (3,-2)

\displaystyle (2, -3)

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

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

Correct answer:

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

Explanation:

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

\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:

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

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

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

Example Question #1 : Midpoint Formula

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Write the midpoint formula.

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

Substitute the points.

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

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

Example Question #1 : Other Coordinate Geometry

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

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

Possible Answers:

\displaystyle 69^{\circ }

\displaystyle 71^{\circ }

\displaystyle 19 ^{\circ }

\displaystyle 21^{\circ }

\displaystyle 34 ^{\circ }

Correct answer:

\displaystyle 19 ^{\circ }

Explanation:

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

\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 

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

using the known points.

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

\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

\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 \displaystyle x_{1}= y_{1} = 0, x_{2}= 3, y_{2} = 2:

\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

\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:

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

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

Example Question #11 : Coordinate Geometry

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

1

Possible Answers:

\displaystyle 3.46

\displaystyle 4.09

\displaystyle 3.99

\displaystyle 3.38

Correct answer:

\displaystyle 3.46

Explanation:

1

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

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

13

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 \displaystyle 1:\sqrt3:2. Thus, the y-coordinate of \displaystyle C must be \displaystyle 3.46.

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