This exam was adminstered in June 2022.
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June 2022 Geometry Regents
Part I
Each correct answer will receive 2 credits. No partial credit.
17. The diagram below shows a tree growing vertically on a hillside. The
angle formed by the tree trunk and the hillside is 100°. The distance
from the base of the tree to the bottom of the hill is 140 feet.
What is the vertical drop, x, to the base of the hill, to the nearest foot?
(1) 24
(2) 25
(3) 70
(4) 138
Answer: (1) 24
The angle formed by the tree and the hill is exterior to the right triangle. The internal angle is 80 degrees. You are given the hypotenuse and you need the adjacent side, so you need to use cosine.
Cos 30 = x / 140
x = 140 cos 30 = 24.3, which is approximately 24, which is Choice (1).
Since the altitude is opposite a 10-degree angle, it must be the shortest side of the triangle, so 138 is impossible and 70 was pretty unlikely on the face of it.
18. On the set of axes below, △LET and △L"E"T" are graphed in the
coordinate plane where △LET ≅ △L"E"T".
Which sequence of rigid motions maps △LET onto △L"E"T"?
(1) a reflection over the y-axis followed by a reflection over the
x-axis
(2) a rotation of 180° about the origin
(3) a rotation of 90° counterclockwise about the origin followed by
a reflection over the y-axis
(4) a reflection over the x-axis followed by a rotation of 90°
clockwise about the origin
Answer: (3) a rotation of 90° counterclockwise about the origin followed by a reflection over the y-axis
The first two choices are easily eliminated. In Choice (1), a reflection over the y-axis would put E' at (1,3) and E" at (1,-3), but E" is at (3,-1) instead. In Choice (2), a rotation of 180 degrees about the origin would also but E" at (1,-3). Eliminate both of these choices.
In Choice (3), a 90 degree CCW rotation moves E to E'(-3,-1). Reflection over the y-axis puts E"(3,-1), which is correct. If you check the other two points, they will also line up.
In Choice (4), a reflection puts the image in Quadrant III (from Quadrant II). A 90 degree clockwise rotation will put the image back in Quadrant II, instead of Quadrant IV. Eliminate Choice (4).
Note that if Choice (4) had a counterclockwise rotation, it would have been correct.
19. Diameter ROQ of circle O is extended through Q to point P, and tangent PA is drawn. If mRA = 100°, what is m∠P?
(1) 10°
(2) 20°
(3) 40°
(4) 50°
Answer: (1) 10°
If you can't remember all the rules about arcs, then think about right triangles.
Make a sketch of the problem. Then add radius OA. Angle ROA = 100 because it's a central angle that intercepts arc RA. This means that AOP is 80 degrees. However, OAP is a right angle because of the tangent line. So 80 + 90 + m∠P = 180, so m∠P = 10.
Using the rules for arcs, m∠P = (mRA - mAQ) / 2 = (100 - 80) / 2 = 20/2 = 10.
20. Segment JM has endpoints J(-5,1) and M(7,-9). An equation of
the perpendicular bisector of JM is
(1) y - 4 = 5/6 (x + 1)
(2) y + 4 = 5/6 (x - 1)
(3) y - 4 = 6/5 (x + 1)
(4) y + 4 = 6/5 (x - 1)
Answer: (4) y + 4 = 6/5 (x - 1)
To find the equationof the perpendicular bisector, you need to know the midpoint of this segment, and the slope of the segment (which you'll use to find the perpendicular slope.)
You need to find the x-coordinate that is in the middle of the two given x-coordinates, and the y-coordinates that is in the middle of the two given y-coordinates. In other words, AVERAGE them. (Add them and divide by 2.)
Midpoint: ( (-5 + 7)/2, (1 + -9)/2 ) = (1, -4)
Note that the equations in the choice are written in Point-Slope form, which is:
So Choice (1) and (3) can be eliminated because they have incorrect signs.
The slope of JM is (-9 - 1) / (7 - -5) = -10/12 = -5/6. The slope of a line perpendicular to a slope of -5/6 is +6/5, which is the inverse reciprocal. (The two slopes have a product of -1.)
Choice (4) is correct.
21. Quadrilateral EBCF and AD are drawn below, such that ABCD is a parallelogram, EB ≅ FB, and EF ⊥ FH.
If m∠E = 62° and m∠C = 51°, what is m∠FHB?
(1) 79°
(2) 76°
(3) 73°
(4) 62°
Answer: (1) 79°
There's a lot going on here, so look at the image below:
The m∠E = 62° and EBF is an isosceles triangle with E as a base angle. This makes m∠EFB = 62° as well. And then m∠EBF = 180 - 2(62) = 56°.
Since EF ⊥ FH, then m∠EFH = 90°. This means that m∠BFH = 90° - 62° = 28°.
ABCD is a parallelogram. Since the m∠C = 51° then the m∠BAD = 51° and the m∠D = 180 - 51° = 129°. So m∠ABC = 129°.
Since m∠ABC = 129° and m∠ABF = 56° then m∠FBH = 129° - 56° = 73°. (We're not done. This is not the angle we're looking for.)
Triangle EFH has angles of 73° and 28°. That is a total of 101°. That means that m∠FBH = 180 - 101 = 79°. This is Choice (1).
22. Point P divides the directed line segment from point A(-4,-1) to point B(6,4) in the ratio 2:3. The coordinates of point P are
(1) (-1,1)
(2) (0,1)
(3) (1,0)
(4) (2,2)
Answer: (2) (0,1)
The ratio 2:3 means that point P is 2/5 of the way from A to B. Unlike the midpoint which is the number in the middle (the halfway point), we need to find a 2/5 point.
The difference between the x values is 6 - -4 = 10. Multiply (10)(2/5) = 4. The x-coordinate will increase 4 units, and -4 + 4 = 0. The only possibility is Choice (2). Let's double-check with y.
The difference between the y values is 4 - -1 = 5, and (5)(2/5) = 2. The y-coordinate will be -1 + 2 = 1. So P is at (0,1), which is Choice (2).
If you went from B to A instead of from A to B, you would have gotten Choice (4) in error, because (6-4,4-2) is (2,2). This is incorrect.
23. A line is dilated by a scale factor of 1/3 centered at a point on the line. Which statement is correct about the image of the line?
(1) Its slope is changed by a scale factor of 1/3.
(2) Its y-intercept is changed by a scale factor of 1/3.
(3) Its slope and y-intercept are changed by a scale factor of 1/3.
(4) The image of the line and the pre-image are the same line.
Answer: (4) The image of the line and the pre-image are the same line.
If a line is dilated centered on a point such as the origina, then all of the points on the line are going to move 1/3 of the distance to the origin. This would NOT change the slope of the line. The line would either be parallel to the original line or it would be the same as the original line.
In this case, the dilation is centered on a point on the line. Every point will come closer to the point of dilation. But a line is infinte, and continues to stretched out in both directions. The line will remain unchanged. So the image and the preimage are the same line. This is Choice (4).
Since the slope would not change, Choice (1) and (3) should have immediately been eliminated.
16.In the diagram below of circle O, tangent AB is drawn from external point B, and secant BCOE and diameter AOD are drawn.
If m∠OBA = 36° and OC = 10, what is the area of shaded sector DOE?
(1) 3π/10
(2) 3π
(3) 10π
(4) 15π
Answer: (4) 15π
If m∠OBA = 36° and m∠OAB = 90° (tangent line), then m∠AOB = 90 - 36 = 54°.
The shaded sector is 54/360 of the circle, which simplifies to 3/20.
The area of the shaded sector then is A = 3/20 π r2, with r = 10.
So A = 3/20 π 102 r = 300/20 π = 15π. This is Choice (4).
End of Part I.
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