This exam was adminstered in January 2023.

More Regents problems.

__January 2023 Geometry Regents__

__January 2023 Geometry Regents__

__Part I __

__Part I__

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

*17. In the diagram below of circle O, AC and BC are chords, and
m∠C = 70°.
*

If OA = 9, the area of the shaded sector AOB is

(1) 3.5π

(2) 7π

(3) 15.75π

(4) 31.5π

If OA = 9, the area of the shaded sector AOB is

(1) 3.5π

(2) 7π

(3) 15.75π

(4) 31.5π

**Answer: (4) 31.5π **

The area of the sector is equal to the fraction of the circle that the sector represents times the area of the entire circle. The fraction is determined by the central angle.

The inscribed angle is half the size of the central angle, so the central angle is 70 times 2 = 140 degrees. This means that the fraction of the circle in the sector is 140/360.

Therefore, the area of the sector is

A = (140/360)(9)^{2}π = 31.5π

This is Choice (4).

*18. Quadrilateral BEST has diagonals that intersect at point D. Which
statement would not be sufficient to prove quadrilateral BEST is a
parallelogram?
(1) BD ≅ SD and ED ≅ TD
(2) BE ≅ ST and ES ≅ TB
(3) ES ≅TB and BE || TS
(4) ES || BT and BE || TS
*

**Answer: (3) ES ≅TB and BE || TS **

There are several ways to show that that a quadrilateral is a parallelogram.

The opposite sides are parallel. This is Choice (4). Eliminate it.

The opposite sides are congruent. This is Choice (2). Eliminate it.

The diagonals bisect each other, which means that each one is split into two congruent segments. This is Choice (1). Eliminate it.

One pair of sides is BOTH parallel and congruent. This is NOT Choice (3). Choice (3) says that one pair of sides is congruent but a *different* pair is parallel. This could be true in an isosceles trapezoid. So Choice (3) is the correct answer.

*19. The equation of line t is 3x - y = 6. Line m is the image of line t
after a dilation with a scale factor of 1/2 centered at the origin.
What is an equation of line m?
(1) y = 3/2 x - 3
(2) y = 3/2 x - 6
(3) y = 3x + 3
(4) y = 3x - 3
*

**Answer: (4) y = 3x - 3 **

Rewrite the equation for line t in slope-intercept form. If m is 1/2 of the original, then the y-intercept of the image is half the distance from the origin. The slope remains unchanged.

3x - y = 6

- y = -3x + 6

y = 3x - 6

Half of -6 is -3, so the equation of m is y = 3x - 3, which is Choice (4).

*20. A cylindrical pool has a diameter of 16 feet and height of 4 feet.
The pool is filled to 1/2 foot below the top. How much water does the
pool contain, to the nearest gallon? [1 ft ^{3} = 7.48 gallons]
(1) 704
(2) 804
(3) 5264
(4) 6016
*

**Answer: (3) 5264 **

If the pool is 4 feet high and is filled to 1/2 foot below then top then it is the height of the water is 3.5 feet. The number of gallons will be the Volume in cubic feet times 7.48 gallons.

Remember that the radius is HALF of the diameter.

The Volume of a cylinder is V = π r^{2} h

V = π (8)^{2} (3.5) = 703.7167...

Multiply 703.717 * 7.48 = 5263.80316, or 5264, which is Choice (3).

*21.The area of △TAP is 36 cm ^{2}. A second triangle, JOE, is formed by
connecting the midpoints of each side of △TAP. What is the area of
△JOE, in square centimeters?
(1) 9
(2) 12
(3) 18
(4) 27
*

**Answer: (1) 9 **

By connecting the midsegments of the triangle, you will divide it into four congruent triangles. So the area will be one quarter of TAP.

One quarter of 36 is 9. This is Choice (1).

*22. On the set of axes below, the endpoints of AB have coordinates A(-3,4) and B(5,2).
*

If AB is dilated by a scale factor of 2 centered at (3,5), what are the coordinates of the endpoints of its image, A'B'?

(1) A'(-7,5) and B'(9,1)

(2) A'(-1,6) and B'(7,4)

(3) A'(-6,8) and B'(10,4)

(4) A'(-9,3) and B'(7,-1)

If AB is dilated by a scale factor of 2 centered at (3,5), what are the coordinates of the endpoints of its image, A'B'?

(1) A'(-7,5) and B'(9,1)

(2) A'(-1,6) and B'(7,4)

(3) A'(-6,8) and B'(10,4)

(4) A'(-9,3) and B'(7,-1)

**Answer: (4) A'(-9,3) and B'(7,-1) **

Points A and B will be moved twice as far from the point (3,5) then they already are.

B is 2 units to the right and 3 units below (3,5). If you double that distance then B' will be located at (5+2, 2-3), which is (7,-1). This is Choice (4).

A is 6 units to the left and 1 unit below (3,5). If you double that distance then A' will be located at (-3-6, 4-1), which is (-9,3). This is still Choice (4), and now we've double-checked it.

*23. In the circle below, AD, AC, BC, and DC are chords, EDF is tangent at point D, and AD || BC.
*

Which statement is always true?

(1) ∠ADE ≅ ∠CAD

(2) ∠CDF ≅ ∠ACB

(3) ∠BCA ≅ ∠DCA

(4) ∠ADC ≅ ∠ADE

Which statement is always true?

(1) ∠ADE ≅ ∠CAD

(2) ∠CDF ≅ ∠ACB

(3) ∠BCA ≅ ∠DCA

(4) ∠ADC ≅ ∠ADE

**Answer: (2) ∠CDF ≅ ∠ACB **

Since AD || BC, then arcs AB and CD are congruent. Angles ACB is half the size of arc AB, and CDF is half the size of arc CD.

Since they are both half the size of congruent arcs then the angles are congruent.

Choice (2) is the correct answer.

Choice (1) ∠ADE ≅ ∠CAD would only work if AC || EF, but would not always be true.

Choices (3) and (4) cannot always be true. All you need to do is slide point A along the circle and you can see how the relative sizes of the listed angles will change.

*24. In the diagram below of △ABC, D and E are the midpoints of AB and AC, respectively, and DE is drawn.
*

I. AA similarity

II. SSS similarity

III. SAS similarity

II. SSS similarity

III. SAS similarity

*Which methods could be used to prove △ABC ~ △ADE?
(1) I and II, only
(2) II and III, only
(3) I and III, only
(4) I, II, and III
*

**Answer: (4) I, II, and III **

DE is a midsegment of triangle ABC. It is parallel to BC and 1/2 the size of BC.

Angle A is congruent to itself through the Reflexive Property.

Angle B is congruent to angle ADE, and angle C is congruent to angle AED because of corresponding angles when transversals cross parallel lines.

Therefore AA similarlity can be used to prove the triangles are similar.

Because D and E are midpoints and angle A is congruent to itself, SAS can be used to prove that they are similar.

And because the midsegment is half the size of BC, SSS similarity can be used to prove that they are similar.

So the answer is I, II, and III, which is Choice (4).

More to come. Comments and questions welcome.

More Regents problems.