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 I: Each correct answer will receive 2 credits.
25. What is the measure of each interior angle of a regular hexagon?
1) 60°
2) 120°
3) 135°
4) 270°
Answer: 2) 120°
Interior angles of a regular polygon could MUST be LESS THAN 180 degrees. Eliminate Choice (4).
As you encounter problems such as this, you'll find yourself "just knowing" many of the answers are surely as you know that a triangle has 180 degrees and a square has 360.
If you keep it going, then a pentagon has 540 and a hexagon has 720. If a hexagon has six angles, each on would be 720/6 = 120.
Choice (1), 60, would be a triangle, and Choice (3) 135, would be an octagon.
26. Which equation represents the perpendicular bisector of AB whose endpoints are A(8,2) and B(0,6)?
1) y = 2x - 4
2) y = -1/2 x + 2
3) y = -1/2 x + 6
4) y = 2x - 12
Answer: 1) y = 2x - 4
You need to know the slope of this line to find what is perpendicular, and you need to know the midpoint.
Except that there's a clue in the choices. To go from (0,6) to (8,2), you have to have a negative slope. ANd that means that the perpendicular bisector will have a positive slope. So you can eliminate Choices (2) and (3).
The midpoint of (0,6) and (8,2) is ( (0+8)/2, (6+2)/2 ) = (4,4). You can substitute x = 4 in Choices (1) and (4) and see which results in y = 4.
y = 2(4) - 4 = 8 - 4 = 4. This is the correct response.
y = 2(4) - 12 = 8 - 4 = -4. Incorrect.
If you wanted to find the slope of the original line: (2 - 6) / (8 - 0) = -4/8 = -1/2. So a line perpendicular to a line with a slope of -1/2 would have a slope of 2.
27. As shown in the diagram below, a kite needs a vertical and a horizontal
support bar attached at opposite corners. The upper edges of the kite
are 7 inches, the side edges are x inches, and the vertical support bar
is (x + 1) inches.
What is the measure, in inches, of the vertical support bar?
1) 23
2) 24
3) 25
4) 26
Answer: 3) 25
You are given a right triangle, with a hypotenuse of (x + 1) and asked to find the length of the hypotenuse. So don't forget to add one.
If you know Pytagorean Triples, then you should know that triangle measures 7-24-25, and the answer is 25.
If you didn't, you'd have to use Pytagorean Theorem:
72 + x2 = (x + 1)2
49 + x2 = x2 + 2x + 1
49 = 2x + 1
48 = 2x
24 = x
The leg is 24, so the vertical bar, which is x + 1, is 25.
V = 1/2 L * W * H = 1/2 (4)(6)(10) = 120
28. Given three distinct quadrilaterals, a square, a rectangle, and a
rhombus, which quadrilaterals must have perpendicular diagonals?
1) the rhombus, onl
2) the rectangle and the square
3) the rhombus and the square
4) the rectangle, the rhombus, and the square
Answer: 3) the rhombus and the square
When they say distinct, they mean that the rhombus isn't also a square and the rectangle isn't a square. Three different shapes.
However, the square is also a rhombus, by defintion.
Perpendicular diagonals is a property of a rhombus, so the square has it as well. The rectangle will not (unless it is a square, but the question says that it is not).
More to come. Comments and questions welcome.
More Regents problems.
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