Geometry Problems of the Day (Geometry Regents, June 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 II: Each correct answer will receive 2 credits. Partial credit is possible.


29. A right circular cylinder has a height of 7 inches and the base has a diameter of 6 inches. Determine the lateral area, in square inches, of the cylinder in terms of π.

Answer:

The lateral area is the surface area not including the top and bottom. If the cylinder were a soup can, it would be the area of the rectangular label on it.

The length of the rectangle is the circumference of the circle. The width is the height of the cylinder.

Therefore, A = (6π)(7) = 42π

30. Determine, in degrees, the measure of each interior angle of a regular octagon.

Answer:

The formula for the sum of the interior angles for a polygon is (n-2)* 180. To find the size of each individual angle in a regular polygon, you have to divide the total by the number of angles: (n-2)*180/n.

An octagon has 8 sides, so (n-2)*180/n = (8-2)(180)/(8) = (6)(180)/(8) = 135.

Alternatively, the sum of the exterior angles of ANY polygon is 360. Each individual exterior angle is a regular polygon is 360/n. This angle is supplementary to the interior angle.

So 180 - 360/8 = 180 - 45 = 135.

31. Triangle ABC has vertices at A(3,0), B(9,-5), and C(7,-8). Find the length of AC in simplest radical form.

Answer:

Note that point B is irrelevant to the problem. You can make a sketch of the triangle in the graph paper in the back of the booklet, but you still need to show work on the question page.

You can use the Distance Formula or Pythagorean Theorem to solve this.

d = SQRT ( (7 - 3)² + (-8 - 0)² )
= SQRT ( 4² + (-8)² )
= SQRT ( 16 + 64 )
= SQRT (80)
= SQRT ( 2 * 2 * 2 * 2 * 5)
= 2 * 2 * SQRT(5)
= 4 SQRT(5)

More to come. Comments and questions welcome.

More Regents problems.

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