Geometry Problems of the Day (Geometry Regents, August 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, August 2013

Part I: Each correct answer will receive 2 credits.


25. In the diagram below, AC and AD are tangent to circle B at points C and D, respectively, and BC, BD, and BA are drawn.

If AC = 12 and AB = 15, what is the length of BD?

1) 6
2) 9
3) 24
4) 36

Answer: 2) 9

BD is a radius of the circle, so is BC. Therefore, BD = BC. BC is the shorter leg of a right triangle because BC is a radius and AC is a tangent, so angle ACB is a right angle.

If you don't recognize the Pythagorean Triple, use the Pythagorean Theorem to solve this.

12² + x² = 15²
144 + x² = 225
x² = 81
x = 9

26. 6 Triangle ABC shown below is a right triangle with altitude AD drawn to the hypotenuse BC.

If BD = 2 and DC = 10, what is the length of AB?

1) 2 * SQRT(2)
2) 2 * SQRT(5)
3) 2 * SQRT(6)
4) 2 * SQRT(30)

Answer: 3) 2 * SQRT(6)

When you draw an altitude in a Right Triangle from the right angle to the hypotenuse, the result is three similar right triangles. That means that the corresponding sides will be proportional.

AB is the shorter leg of BAC. It is also the hypotenuse of BDA. So you can set up a proportion of shorter leg to hypotenuse:

x / 12 = 2 / x
x² = 24
x = SQRT(24)
x = SQRT(2 * 2 * 2 * 3)
x = 2 * SQRT(2 * 3)
x = 2 * SQRT(6)

27. Triangle ABC has vertices A(0,0), B(6,8), and C(8,4). Which equation represents the perpendicular bisector of BC?

1) y = 2x - 6
2) y = -2x + 4
3) y = 1/2 x + 5/2
4) y = -1/2 x + 19/2

Answer: 3) y = 1/2 x + 5/2

The perpedicular bisector of BC will have a slope that is the negative reciprocal of the slope of BC and go through the midpoint of BC.

Note 1: Point A is irrelevant to this problem.

Note 2: All four choices have different slopes, so you don't need to find the midpoint of BC.

The slope of BC is (4 - 8) / (8 - 6) = -4 / 2 = -2. Therefore, the slope of the perpendicular bisector must be +1/2.

28. Chords AB and CD intersect at point E in a circle with center at O. If AE = 8, AB = 20, and DE = 16, what is the length of CE?

1) 6
2) 9
3) 10
4) 12

Answer: (1) 6

If AB intersects CD at point E then the product of (AE)(BE) must be equal to the product of (CE)(DE). Also, if AB = 20 and AE = 8, then BE = 20 - 8 = 12

(AE)(BE) = (CE)(DE)
(8)(12) = (CE)(16)
96 = 16(CE)
6 = (CE)

Just for giggles, another way to solve the above problem:

16(CE) = (8)(12) = (2)(2)(2)(2)(2)(3) = (2)(2)(2)(2) * (2)(3) = (16)(6)
So CE = 6
You can always use those Properties of Real Numbers that you learned in Algebra class.

More to come. Comments and questions welcome.

More Regents problems.

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