Geometry Problems of the Day (Geometry Regents, August 2011)

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, August 2011

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


32. Write an equation of the circle graphed in the diagram below.

Answer:

The equation of a circle is given by the equation (x - h)² + (y - k)² = r², where (h,k) is the center of the circle and r is the radius.

The center of the circle is (5,-4). The radius is 6. (Count from the center to the top or the bottom.)

So the equation is (x - 5)² + (y + 4)² = 36</P>

33. The diagram below shows △ABC, with AEB, ADC, and ∠ACB ≅ ∠AED. Prove that △ABC is similar to △ADE.

Answer:

Two triangles are similar if they have two angles that are congruent. (The third will be automaticallys congruent.)

It is given that ∠ACB ≅ ∠AED, and ∠A ≅ ∠A by the Reflexive property. So △ABC is similar to △ADE by the AA Similarity Theorem.

34. Triangle ABC has vertices A(3,3), B(7,9), and C(11,3). Determine the point of intersection of the medians, and state its coordinates. [The use of the set of axes below is optional.]

Answer:

The use of the set of axes below is optional but this is one time that it is probably better than trying to find the answer algebraically.

Graph the triangle. Then graph the medians from each vertex to the middle of the opposite side. You will only need to draw two of them, but draw the third one as a check.

The point of intersection (point of concurrence) is (7,5).

This one isn't too difficult when you notice that AC is a horizontal line, and its midpoint is (7,3). This means that the median from B to AC is a vertical line with a length of 6. The cetroid will be 2/3 of the way from B to AC, which is 4 units below (7,9). That would be (7,5).

This might have been more complicated if it were a Part IV question.

More to come. Comments and questions welcome.

More Regents problems.

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