Geometry Problems of the Day (Geometry Regents, August 2010)

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 2010

Part II: Each correct answer will receive 2 credits. Partial credit is available. One computational or conceptual error will cost 1 credit. A second will cost the other.


29. The diagram below shows isosceles trapezoid ABCD with AB || DC and AD ≅ BC. If m∠BAD = 2x and m∠BCD = 3x + 5, find m∠BAD.

Answer:

The fact that AB || DC establishes which sides of the trapezoid are the bases. The fact that AD ≅ BC means that it is an isosceles trapezoid.

That means that m∠BAD = m∠ABC = 2x. Since angles B and C are same-side interior angles, m∠B + m∠C = 180.

So 2x + 3x + 5 = 180
5x + 5 = 180
5x = 175
x = 35

If x = 35, the 2x = 70. So m∠BAD = 70.

30. A right circular cone has a base with a radius of 15 cm, a vertical height of 20 cm, and a slant height of 25 cm. Find, in terms of π, the number of square centimeters in the lateral area of the cone.

Answer:

The lateral area of the cone is the surface area of just the slanted portion without the circular base.

The back of the booklet gives the formula for Lateral Area of a Right Circular Cone as L = πrl where l is the slant height

Multiply: L = πrl = π(15)(25) = 375π

That's it. You didn't have to change radius to anything else. You didn't have to find the height of the cone. Just substitute what you were given. DO NOT REPLACE π with 3.14 or any other values.

27. In the diagram below of △HQP, side HP is extended through P to T, m∠QPT = 6x + 20, m∠HQP = x + 40, and m∠PHQ = 4x − 5. Find m∠QPT.

Answer:

The Exterior Angle Theorem tells us that m∠QPT is the sum of the two Remote Angles, ∠PQT and ∠QHP.

6x + 20 = x + 40 + 4x - 5
6x + 20 = 5x + 35
x = 15

m∠QPT = 6(15) + 20 = 110

More to come. Comments and questions welcome.

More Regents problems.

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