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 II: Each correct answer will receive 2 credits. Partial credit is available.


32. The coordinates of the vertices of triangle RST are R(-2,3), S(4,4), and T(2,-2). Triangle R'S'T' is the image of triangle RST after a rotation of 90° about the origin.
State the coordinates of the vertices of Triangle R'S'T'.

Answer:

When a direction is not specified, the rotation is Counter-clockwise, like the numbering of the quadrants.

If you graph the points you will see the rule. The x-coordinate and the y-coordinate will switch places. After they switch, the x-coordinate will change signs.

That is, P(x, y) becomes P'(-y, x).

R(-2,3) -> R'(-3,-2) -- moves from Q II to Q III

S(4,4) -> S'(-4,4) -- moves from Q I to Q II

T(2,-2) -> S'(2,2) -- moves from Q IV to Q I

It you make a note of the quadrants for yourself, you shouldn't mess up the signs of the numbers.

33. On the diagram below, use a compass and straightedge to construct the bisector of ∠XYZ. [Leave all construction marks.]

Answer:

TO bisect an angle, make an arc from the vertex. Then make the same arc from each of the points where the arc intercepts the sides of the angle. Make a line from Y to the intersection of the two arcs.

Look at the illustration below:

34. In the diagram below of circle O, diameter AB is perpendicular to chord CD at E.

If AO = 10 and BE = 4, find the length of CE.

Answer:

Since diameter AB is perpendicular to CD, it also bisects CD. If you draw OC and OD, they will both be radii equal to AO, which is 10. You will also have two right triangles.

OC = 10. OE = OB - BE = 10 - 4 = 6.

At this point, you should already know the answer.

6² + x² = 10²
36 + x² = 100
x² = 64
x = 8

The length of CE = 8.

The slope of the first line is 2. The slope of the second line is -1/2.

Since the two slopes are inverse reciprocals, with a product of -1, so the lines are perpendicular.

End of Part II.

More to come. Comments and questions welcome.

More Regents problems.

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