What follows is a portion of the Common Core Geometry exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.

Part 1 is was posted here.

### August 2016 Geometry Regents, Part II

**25. ***5 Lines AE and BD are tangent to circles O and P at A, E, B, and D, as shown in the diagram below.
If AC:CE = 5:3, and BD = 56, determine and state the length of CD
*

Because they are tangent lines, AC = BC and CD = CE. So if AC:CE = 5:3, then BC:CD = 5:3 also.

So 5x + 3x = 56

8x = 56

x = 7

CD = 3x = 3(7) = 21.

**26. ***6 In the diagram below, triangle ABC has coordinates A(1,1), B(4,1), and C(4,5). Graph and label
triangle A" B" C" , the image of ABC after the translation five units to the right and two units up followed by the reflection over the line y 0.
*

First, remember that the line y = 0 is the x-axis, not the y-axis.

Move each point 5 units to the right and 2 units up. Point A (1, 1) will move to A'(6, 3), etc.

Next, flip the points over the x-axis, so point A'(6, 3) becomes A"(6, -3).

You're final image will look like this.

**27. ***A regular hexagon is rotated in a counterclockwise direction about its center. Determine and state
the minimum number of degrees in the rotation such that the hexagon will coincide with itself.
*

A regular hexagon will coincide with itself every 1/6 of a revolution. (1/6)(360) = 60 degrees.

**28. ***In the diagram of ABC shown below, use a compass and straightedge to construct the median
to AB. [Leave all construction marks.]
*

*Construction marks are difficult for me to show using the tools I'm currently using, so I will describe the process.*

The median is a line drawn from point C to the midpoint of AB, so you need to find the midpoint of AB. The way to do this is to construct the perpendicular bisector of AB, label the midpoint and then draw a line from C to this new point.

The red lines are created by opening the compass so that it is wider than half the distance between A and B. If you make the arcs to small, then they won't intersect. You will have to make the arc bigger and try again. Remember that both arcs MUST be the same size when you center on A and when you center it on B.

Draw the perpendicular bisector, which is the line through the two points where the arcs meet. This line will cross AB at its midpoint.

Draw the median by using the straightedge and marking the line between C and the midpoint.

**29. ***Triangle MNP is the image of triangle JKL after a 120° counterclockwise rotation about point Q.
If the measure of angle L is 47° and the measure of angle N is 57°, determine the measure of
angle M. Explain how you arrived at your answer.
*

When the triangles rotate, M is the image of J, N is the image of K and P is the image of L. Size and shape are preserved in a rotation so <M = <J, <N = <K annd <P = <L.

&LtM + &LtN + <P = 180

<M + 47 + 57 = 180

<M + 104 = 180

<M = 76

**30. ***A circle has a center at (1,–2) and radius of 4. Does the point (3.4,1.2) lie on the circle?
Justify your answer.
*

If (3.4, 1.2) is on the circle then the following equation must be true.

(3.4 - 1)^{2} + (1.2 + 2)^{2} = 4^{2} ?

(2.4)^{2} + (3.2)^{2} = 16 ?

5.76 + 10.24 = 16 ?

16 = 16. Check

(3.4, 1.2) lies on the circle

**31. ***In the diagram below, a window of a house is 15 feet above the ground. A ladder is placed against
the house with its base at an angle of 75° with the ground. Determine and state the length of the ladder to the nearest tenth of a foot.
*

You have a right triangle. You have an angle. You have the opposite side. You want to know the hypotenuse. O-H means you need to use the sine ratio. (Make sure your calculator is in **DEGREE** mode.

sin 75 = 15 / x or as a proportion: sin 75 / 1 = 15 / x.

cross-multiply: (sin 75) x = 15

x = 15 / sin 75 = 15.5291427062

The ladder is 15.5 feet to the nearest tenth.

**END OF PART II**

How did you do? Any questions? (I appreciate pointing out any "typos" in my problems. Thank you.)

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