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

Part I: Each correct answer will receive 2 credits.


25. Which graph represents a circle whose equation is x² + (y - 1)² = 9?

Answer: 1) [SEE IMAGE]

The center of the circle is (0, 1) and the radius is 3. That is Choice (1).

The equation for a circle is (x - h)² + (y - 1)² = r², where (h,k) is the center and r is the radius.

26. What is the perimeter of a rhombus whose diagonals are 16 and 30?

1) 92
2) 68
3) 60
4) 17

Answer: 2) 68

The diagonals of a rhombus will split the quadrilateral into 4 congruent right triangles. Since the diagonals are 16 and 30, then the legs of the right triangle will be 8 and 15.

Using Pythagorean Theorem, the hypotenuse, which is the side of the rhombus, will be 17. (You should know your Pythagorean Triples.)

The perimeter is 4 * 17 = 68.

27. In right triangle ABC shown in the diagram below, altitude BD is drawn to hypotenuse AC, CD = 12, and AD = 3.

What is the length of AB?

1) 5 SQRT(3)
2) 6
3) 3 SQRT(5)
4) 9

Answer: 3) 12

According to the Right Triangle Altitude Theorem, (AD)(DC) = (DB)².

(DB)² = (3)(12) = 36
DB = 6
But this isn't the side we are looking for.

Using Pythagorean Theorem, 3² + 6² = (AB)²
9 + 36 = (AB)²
45 = (AB)²
AB = SQRT(45) = SQRT(9 * 5) = SQRT(9)SQRT(5) = 3 SQRT(5)

More directly, you could have written the proportion:

AB / AC = AD / AB
AB / 15 = 3 / AB

... which would have led to the same answer.

28. Secants JKL and JMN are drawn to circle O from an external point, J. If JK = 8, LK = 4, and JM = 6, what is the length of JN?

1) 16
2) 12
3) 10
4) 8

Answer: 1) 16

The Secant Secant Theorem tells us that the following products are equal: (JK)(JL) = (JM)(JN)

If LK = 4 then JL = 8 + 4 = 12.

(8)(12) = (6)(JN)
96 = 6(JN)
JN = 16

End of Part I.

More to come. Comments and questions welcome.

More Regents problems.

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