Friday, November 19, 2021

Geometry Problems of the Day (Geometry Regents, January 2012)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, January 2012

Part III: Each correct answer will receive 4 credits. Partial credit is available.

35. Triangle ABC has coordinates A(2,-2), B(2,1), and C(4,-2). Triangle A'B'C' is the image of triangle ABC under T5,-2.

On the set of axes below, graph and label triangle ABC and its image, triangle A'B'C'.

Determine the relationship between the area of triangle ABC and the area of tirangle A'B'C'.

Justify your response.


Here is the graph of ABC and its image, which is five units to the right and two units down.

The area of the triangle and its image is the same because a translation is a rigid motion which preserves size, shape and area.

Alternatively, since it the question doesn't say how to justify your response, you could find the area of both triangles:

Area of ABC = 1/2(3)(2) = 3

Area of A'B'C' = 1/2(3)(2) = 3

Area of ABC = Area of A'B'C'.

36. A paint can is in the shape of a right circular cylinder. The volume of the paint can is 600π cubic inches and its altitude is 12 inches.

Find the radius, in inches, of the base of the paint can. Express the answer in simplest radical form.

Find, to the nearest tenth of a square inch, the lateral area of the paint can.


The lateral area of a cylinder is the surface area without the top and bottom, like a tube, or a label on a soup can. If you open up the cylinder, it's a rectangle with a length equal to the circumference of the circles on top and bottom (and width equal to the height of the cylinder).

V = π r2 h

600π = π r2 (12)

r2 = 600 / 12 = 50

r = √(50) = √(25*2) = 5 √(2)

The radius is 5 √(2).

LA = 2 π r h

LA = 2 π 5√(2) (12) = 533.14...

The lateral area is 533.1.

37. Triangle HKL has vertices H(-7,2), K(3,-4), and L(5,4). The midpoint of HL is M and the midpoint of LK is N.

Determine and state the coordinates of points M and N.

Justify the statement: MN is parallel to HK.

[The use of the set of axes below is optional.]


Normally, I wouldn't bother with the graph and do everything algebraically. But when it comes to proveing that two lines are parallel because they have the same slope, it's nice to it visually. Also, when you see the boxes in front of you, it's harder to make a sign error that dives you the wrong coordinate.

MN is parallel to HK because they have the same slope as shown in this graph:


Midpoint of HL is M( (-7+5)/2, (2+4)/2 ) = M(-1,3)

Midpoint of LK is N( (3+5)/2, (-4+4)/2 ) = N(4,0)

To show that MN is parallel to HK, find the slopes of each line:

Slope of MN = (3 - 0) / (-1 - 4) = -3/5.

Slope of HK = (-4 - 2) / (3 - -7) = -6/10 = -3/5.

HK and MN have the same slope, so they are parallel.

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

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