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 III: Each correct answer will receive 4 credits. Partial credit is possible.
35. On the set of axes below, graph the locus of points that are four units from the point (2,1). On the same set of axes, graph the locus of points that are two units from the line x = 4. State the
coordinates of all points that satisfy both conditions.
Answer:
The first locu of points is a circle with a radius of 4 and a center at (2,1). The second locus of points are two vertical lines, x = 2 and x = 6. The first line intersects the circle in two places. The second line is tangent to the circle.
The points that satisfy both conditions are (2,5), (2,-3), and (6,1).
See the image below:
36. In the diagram below, BFCE, AB ⊥ BE, DE ⊥ BE, and ∠BFD ≅ ∠ECA.
Prove that △ABC ∼ △DEF.
Answer:
To prove that two triangles are similar, you have to show that two sets of angles are congruent. You are given one pair of angles and also that the pairs of sides are perpendicular, meaning that each triangle has a right angle. That's two angles, so you can use AA.
Your proof would go something like this:
Statement | Reason |
1. AB ⊥ BE, DE ⊥ BE, ∠BFD ≅ ∠ECA | 1. Given |
2. Angle B is a right angle. | 2. Definition of perpendicular. |
3. Angle E is a right angle. | 3. Definition of perpendicular. |
4. ∠B ≅ ∠E | 4. All right angles are congruent. |
5. △ABC ∼ △DEFE | 5. AA Theorem |
37. In the diagram below of △ADE, B is a point on AE and C is a point on AD such that BC || ED,
AC = x − 3, BE = 20, AB = 16, and AD = 2x + 2. Find the length of AC.
Answer:
The triangles are similar so the sides are proportional. AB/AC = AE/AD.
(16)/(x - 3) = (16 + 20) / (2x + 2)
16(2x + 2) = 36(x - 3)
32x + 32 = 36x - 108
140 = 4x
x = 35
AC = x - 3 = 35 - 3 = 32.
End of Part III.
More to come. Comments and questions welcome.
More Regents problems.
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