Monday, September 20, 2021

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.


32. On the ray drawn below, using a compass and straightedge, construct an equilateral triangle with a vertex at R. The length of a side of the triangle must be equal to a length of the diagonal of rectangle ABCD.


Answer:


Use the straightedge to draw diagonal AC. This isn't essential but it's a good visual. Put the compass at point A and open it to the length of AC. Make a little arc through point C.

Next, without changing the compass, move it to point R and make and arc from the ray counterclockwise above the ray. Then move the compass to the point where the arc intercepted the ray and make another arc above the ray that intersects the other one. The point of intersection is the third point of the equilateral triangle.

Use the straightedge to complete the triangle.





33. On the set of axes below, graph the locus of points 4 units from the x-axis and equidistant from the points whose coordinates are (-2,0) and (8,0). Mark with an X all points that satisfy both conditions.

Answer:


The locus of points 4 units from the x-axis are the two horizontal parallel lines y = 4 and y = -4. The points equidistant from (-2,0) and (8,0) are sit on the perpendicular bisector of a line segment joining those points.

First draw y = 4 and y = 4.

Next, both (-2, 0) and (8, 0) are on the x-axis, which is a horizontal line. The midpoint between those two points is ( (8-2)/2, 0), or (3, 0). Draw a vertical line at x = 3.

Place an X at point (3, 4) and (3, -4).





34. The coordinates of two vertices of square ABCD are A(2,1) and B(4,4). Determine the slope of side BC.

Answer:


Squares have right angles, so the sides are perpendicular. Find the slope of AB. The slope of BC is the

The slope of AB is (4 - 1) / (4 - 2) = 3 / 2.

The slope of BC is -2/3 because that is the inverse reciprocal, and (3/2)(-2/3) = -1.




End of Part II.

More to come. Comments and questions welcome.

More Regents problems.

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