(x, why?): Geometry Problems of the Day (Geometry Regents, January 2025 Part I)

This exam was adminstered in January 2025.

January 2025 Geometry Regents

Part I

Each correct answer will receive 2 credits. No partial credit.

9. Scalene triangle JKL is drawn below.

If median LM is drawn to side KJ, which statement is always true?

(1) LM = KM
(2) KM = 1/2 KJ
(3) LM ⊥ KJ
(4) ∠KLM ≅ ∠JLM

Answer: (2) KM = 1/2 KJ

The median connects angle L with the midpoint of KJ. Therefore, KM is half of the length of KJ.

LM is not congruent to KM because LM is the median. KM is congruent to JM, but they are each half of KJ.

10. In circle O, chord KA intersects diameter YN at S.

If mYK = 120° and mYA = 105°, what is m∠ASN?

(1) 22.5°
(2) 75°
(3) 97.5°
(4) 120°

Answer: (3) 97.5°

Since YN is a diameter, we can find the measures of the other two arcs. Once we know those, the measure of angle ASN will be the average of the measures of arcs YK and AN.

If YK = 120 degrees, then KN = 60 degrees. If YA = 105 degrees, then AN = 75 degrees.

The angle ASN will be the average, half the sum, of the measures of arcs YK and AN. So 1/2(120 + 75) = 1/2(195) = 97.5, which is Choice (3).

11.Triangle ABC is graphed on the set of axes below. The vertices of △ABC have coordinates A(-3,4), B(-5,-1), and C(3,-2).

What is the area of △ABC?

(1) 16
(2) 20
(3) 21
(4) 24

Answer: (3) 21

The easiest way to find the area of a triangle that does line up with the grid is to create a box around it and then find the area of the three triangles that you cut away from the rectangle.

If you make a box from point C(3,-2) to (3,4) to (-5,4) to (-5,-2), you have a rectangle with an area of 8 x 6 = 48 square units. The three extraneous triangles will have areas of 1/2(2)(5) = 5, 1/2(1)(8) = 4, and 1/2(6)(6) = 18. Subtract 48 - 5 - 4 - 18 = 21, which is Choice (3).

In this example, it would've been possible to divide the triangle along the line y = -1. In that case, the top triangle would have an area of 1/2(7)(5) = 17.5, and the bottom triangle would have an area of 1/2(7)(1) = 3.5. The sum of 17.5 + 3.5 = 21.

12. In △ABC below, DE is a midsegment, and BD ≅ DE.
Which statement is always true?

(1) △ABC is isosceles
(2) △ABC is scalene
(3) BD ≅ BE
(4) DA ≅ EC

Answer: △ABC is isosceles

If BD ≅ DE, then BA ≅ AC because the latter two line segments are twice the size of the first two line segments. If you double the size of congruent segments, you will get another pair of congruent segments that are twice the size.

That means that ABC is an isosceles triangle because two sides, BA and AC, are congruent.

Even though the image shown appears to be scalene, there is no reason that the triangle described in the question must be scalene. You cannot assume that the image shown is drawn to scale and covers all possibilities.

There is nothing indicating that BD must be congruent with BE. Moreover, if BD were congruent to BE, then DA would have to be congruent to EC as well. They can't both be true.

13. As shown in the diagram below, JKL || MNOP, KRN, and OR ≅ ON.

If m∠POR = 116°, what is m∠LKN?

(1) 58°
(2) 116°
(3) 122°
(4) 128°

Answer: (4) 128°

Work your way through the angles you know and the ones you can work out.

If m∠POR = 116°,then m∠NOR = 64° because they are a linear pair. If m∠NOR = 64°, then m∠NRO = 64° because NOR is an isosceles triangle and OR ≅ ON. Then m∠ONR = 52° because 180 - 64 - 64 = 52 degrees.

Since JKL || MNOP, then ∠LKN and ∠PNK are supplementary because they are same-side interior angles. Therefore m∠LKN = 180 - 52 = 128, which is Choice (4).

14. The ratio of similarity of square ABCD to square WXYZ is 2:5. If AB = x + 3 and WX = 3x + 5, then the perimeter of ABCD is

(1) 8
(2) 20
(3) 32
(4) 80

Answer: (3) 32

Solve for x using the lengths of AB and WX and the ratio between them. Once you know the length of one side, you can find the perimeter of the square.

5(x + 3) = 2(3x + 5)
5x + 15 = 6x + 10
5 = x

AB = 5 + 3 = 8
Perimeter of ABCD is 4 * 8 = 32.

The correct answer is Choice (3).

15. In parallelogram ABCD below, diagonals AC and BD intersect at E.

Which transformation would map △ABC onto △CDA?

(1) a reflection over AC
(2) a reflection over DB
(3) a clockwise rotation of 90° about point E
(4) a clockwise rotation of 180° about point E

Answer: (4) a clockwise rotation of 180° about point E

A reflection over the diagonals would not line up correctly because neither AC nor BD are angle bisectors. The answer must be a rotation.

To move A to C, the triangle would have to be rotated 180 degrees about point E. That's Choice (4), which is the correct answer.

16. The square pyramid drawn below has a volume of 175.

If the height of the pyramid is 21, what is the perimeter of the base?

(1) 5
(2) 10
(3) 20
(4) 25

Answer: (3) 20

If the Volume is 175 and the height is 21, then the area of the base can be found using the Volume formula:

V = 1/3 (Area of Base) (height)
175 = 1/3 (A) (21)
175 = 7 A
25 = A

If the Area of the square is 25, then the length of one side is the square root of 25, which is 5. If the length of one side is 5, then the Perimeter of the square base is 4 * 5 = 20, which is CHoice (3).

More to come. Comments and questions welcome.

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