This exam was adminstered in June 2025.
June 2025 Geometry, Part II
Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.
25. In the year 2020, the village of Depew, New York had an area of 5.1 square miles and a population of 15,069. In the same year, the village of Lancaster, New York had an area 2.7 square miles and a population of 10,087.
Which village had the larger populatio density in 2020? Justify your answer.
Answer:
Density is mass divided by volume, but population density is population divided by square footage. Find the density of both cities.
The population density of Depew is 15069/5.1 or about 2954.7.
The population density of Lancaster is 10087/2.7 or about 3735.9.
Lancaster had the larger population density.
You only got both credits if you answered the question and showed work. An answer of "Lancaster" without any work is woth 0 points.
26. In △ABC below, AC is extended through C to D, m∠A = (3x - 22)o, m∠A = (4x - 18)o, and m∠BCD = (6x - 23)o.
Determine and state m∠ACB.
Answer:
This is a two-part problem and, if you ask me (and no one did), a little too involved for a 2-point question.
The exterior angle is equal to the sum of the two remote angles. You need to write an equation to solve for x. Then you have to find the size of angle BCD. AND THEN you have to find the measure of the angle supplementary to the exterior angle.
Again, a little much.
3x - 22 + 4x - 18 = 6x - 23
7x - 40 = 6x - 23
x = 17
This means that the exterior angle is 6(17) - 23 = 79. So angle ACB = 180 - 79 = 101 degrees.
27. Parallelogram ABCD is shown below. Using a compass and straightedge, construct the altitude from point A to side DC.
[Leave all construction marks.]
Answer:
From point A, draw an arc the length of AD. OR extend line CD and draw an arc of any length that intersects CD in two places.
From the two points of line CD, make arcs below CD. Make a point where the two arcs intersect. Call it anything, say E. with a straightedge, draw AE.
That's it.
If you made the altitude from points B, C, or D, you only got one point. If you did it from any other point, or if you made a perpendicular bisector of CD, you score no points.
28. Quadrilateral QUAD is graphed on the set of axes below.
Determine and state the area of quadrilateral QUAD.
Answer:
It was quite easy to show that the diagonals of the quadrilater were perpendicular to each other because the diagonals were vertical and horizontal lines. However, the question did not say that this was a parallelogram. It only said it was a quadrilateral.
It is also easy to see that QUAD could be broken up into four congruent triangles, for which the area could be found.
HOWEVER, this past January, there was a similar question where the figure was NOT so nice. As a result, the easiest way to find the area was to draw a rectangle box around the figure, find the area of individual triangles between the shape and the rectangle, and then subtract. What remained was the area of the figure. That wasn't necessary to do her, BUT anyone who practiced that exam in the days before this exam would've been more likely to take this approach.
I will say that I saw a few answers based on the fact that it is a rhombus. To use those formulas, however, you needed to show that it was a rhombus. This was a lot of extra work for 2 points, but more power to whoever did it this way!
If you draw the two diagonals, you will split QUAD into four triangles. Each of those triangles has a base of 5 and a height of 3.
So Area = 4 (1/2) (3) (5) = 30
If you found the area of the rectangle around QUAD, it had a length of 10 and a width of 6, for an Area of 60. If you then subtracted the four triangles, each of which was 5 by 3, you got 60 - 30 = 30.
Common mistakes were finding the perimter, or finding the lengths of the sides and then sqauring it, assuming that the shape was a square.
29. In a right triangle, the acute angles have the relationship sin(3x - 7)o = cos(x + 1)o
Determine and state the value of x.
Answer:
If the sine value is equalt to the cosine value than the sum of the two angles must be 90 degrees. (The "co" is "cosine" means "complementary", and complementray angles add up to 90 degrees.
3x + 7 + x + 1 = 90
4x - 6 = 90
4x = 96
x = 24
Common mistakes are to set the two expressions equal to each other, which is a conceptual error, setting the sum equal to 180, another conceptual error, or any computational mistakes solving the algebraic equation.
One mistake costs one point. Two points results in a score of zero.
30. In circle A below, m∠BAM = 36°
If AB = 20, determine and state the length of arc MB. [Leave your answer in terms of π.]
Answer:
The circumference of a full circle is 2πr. For an arc, it is the circumference times the fraction of the circle, which is denoted by the measure of the circle arc divided by 360 degrees.
So 36/360 (2)(20)π = 4π.
That's all there was to this.
31. In triagnels ANT and ELM below, AN = 6, NT = 5.6, TA = 4, EL = 9, LM = 8.4, and ME = 6.
Explain why △ANT ~ △ELM
Answer: You can show that two triangles are similar by showing that all three pairs of corresponding sides are proportional by SSS. Also, one triangle is the image of the other after a dilation of a particular scale factor. ("Obviously", saying this is not enough. You need to provide the numbers.)
6/4 = 1.5, 9/6 = 1.5, and 8.4/5.6 = 1.5.
Triangle ELM is a dilation of ANT by a scale faction of 1.5. All of the corresponding sides are proportional so the two are similar by SSS.
Note: if SSS wasn't mentioned, you likely lost a point.
End of Part II
How did you do?
Questions, comments and corrections welcome.
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