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 2011
Part II: Each correct answer will receive 2 credits. Partial credit is possible.
32. In the diagram below of circle O, chord AB bisects chord CD at E. If AE = 8 and BE = 9,
find the length of CE in simplest radical form.
Answer:
According to the Intersection Chords Theorem, the product (AE)(BE) = (CE)(DE). Since CD is bisected, then CE = DE.
So x2 = (8)(9) = 72
Then x = √(72) = √(36 * 2) = 6√(2)
30. On the diagram below, use a compass and straightedge to construct the bisector of ∠ABC.
[Leave all construction marks.]
Answer:
To bisect the angle, place the compass at point B. Draw an arc from BA to BC. From the points on the rays where the arc intercepted it, draw another arc. These two arcs will intersect each other. Mark a point and then draw the bisector from B through this point.
See the image below:
34. Find the slope of a line perpendicular to the line whose equation is 2y − 6x = 4.
Answer:
The slopeof the line perpendicular to the given line will have a slope that is the inverse reciprocal. That is, the two slopes will have a product of -1.
The equation in Standard Form is -6x + 2y = 4. The slope of that line is -A/B = -(-6)/2 = 3.
The slope of the perpendicular line would therefore be -1/3.
End of Part II.
More to come. Comments and questions welcome.
More Regents problems.
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