Wednesday, October 06, 2021

Geometry Problems of the Day (Geometry Regents, January 2013)

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, June 2013

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

35. 5 A right circular cylinder with a height of 5 cm has a base with a diameter of 6 cm. Find the lateral area of the cylinder to the nearest hundredth of a square centimeter.
Find the volume of the cylinder to the nearest hundredth of a cubic centimeter.


The lateral area of a cylinder is the surface area without the top and bottom. If the cylinder were a soup can, then the lateral area would be the area of the label around the can.

The lateral area is found by multiplying the length times the width of the rectangle that is wrapped around the cylinder. The width is the height of the cylinder, and the length is the circumference of the base.

So LA = π d h = (π)(6)(5) = 30 π = 30(3.141592...) = 94.25.
Note that if you used 3.14 for pi, you got an incorrect answer.

Volume = π r2 h = (3.141592...)(3)2(5) = 141.37164... = 141.37.

36. Triangle ABC has vertices A(5,1), B(1,4), and C(1,1). State and label the coordinates of the vertices of triangle A"B"C", the image of triangle ABC, following the composite transformation T1,–1 ° D2. [The use of the set of axes below is optional.]


It is important to remember that in a Composition of Transformations, that dot is the middle can be read as the word "OF". You want the Translation OF THE Dilation, not a Translation followed by a Dilation.

To perform a Dilation of scale factor 2, multiply all of the coordinates by 2:

A(5, 1) --> A'(10, 2)
B(1, 4) --> B'(2, 8)
C(1, 1) --> C'(2,2)

To perform a Translation of (x+1, y-1), add 1 to each x-coordinate and subtract 1 from each y-coordinate.

A'(10, 2) --> A"(11, 1)
B'(2, 8) --> B"(3, 7)
C'(2, 2) --> C"(3,1)

If you use the graph, make sure to label the points with the correct coordinates.

37. In triangle ABC, m∠A = x2 + 12, m∠B = 11x + 5, and m∠C = 13x - 17.
Determine the longest side of triangle ABC.


The longest side will be the same across from the biggest angle. The sum of the three angles must be 180 degrees. Write and solve a quadratic equation to find x and then use that value to find the size of each individual angle.

x2 + 12 + 11x + 5 + 13x - 17 = 180
x2 + 24x + 0 = 180
x2 + 24x - 180 = 0
(x + 30)(x - 6) = 0
x + 30 = 0 or x - 6 = 0
x = - 30 or x = 6
Discard the negative value.

Angle A = (6)2 + 12 = 48. Angle B = 11(6) + 5 = 71. Angle C = 13(6) - 17 = 61.

Since Angle B is the largest angle, then AC is the longest side of the triangle.

Note that a guess of AC without any work is worth 0 credit. Find the sizes of the three angles without answering the question is worth 3 credits.

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

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