Thursday, January 30, 2014

January 2014 Geometry Regents Discussion Thread, Part 3: Multiple Choice

Previous threads: Part 1, Part 2.

I'm not going to type up all the questions, but I'll identify the topic addressed in the problems.

1. You are given a midpoint and an endpoint of a line segment and have to find the other endpoint. You don't really need to use the midpoint formula and solve algebraically. Just notice that from A to M the x-co-ordinate drops 2. Drop it two more and B has an x-co-ordinate of 2. There's only one choice that works, so it's the answer, and you don't have to find y, unless you want to check.

2. Which construction is used to make a 45o angle? Only two choices have 45o angles in them (or something that looks like one). Choice (3) shows a straight angle bisected and then the right angle bisected.

3. First mention of the equation of a circle: (x - h)2 + (y - k)2 = r2
If there's a "+" in the equation before h or k, it means that the co-ordinate is negative. Take the square root of the number after the equal sign.

4. If two planes are perpendicular to the same line then they are parallel to each other.

5. A triangle is transformed and the image is congruent to the original. This is true from reflections, translation and rotations but not dilations, which will expand or shrink the size of the triangle.

6. For two edges of a cube not to be coplanar, they have to point in different directions; i.e., skew lines. Remember: two edges don't have to be on the same surface of the prism for them to be coplanar.

7. The points that are 3 units away from the origin and equidistant from the x-axis and y-axis would be on the lines y = x and y = -x where they intersect the circle centered on the origin with a radius of three. Therefore, there are four points which fit both conditions.

8. The medians of a triangle meet at the centroid. The longer portion, from the vertex to the centroid, is twice the length of the other portion, from the centroid to the opposite side of the triangle. That makes the shorter piece one-third the length of the median, and the longer piece two-thirds. If the median is 12, the segments are 8 and 4.

9. Solve y = x and y = x2 - 2.
You could do substitution and factoring. Or you could just try (1, 1) and (2, 2). (1, 1) doesn't work. (2, 2) does. That leaves choice (2) (2,2) and (-1, -1). A quick double-check on (-1, -1) shows that it's the answer.

10. Parallel lines have the same slope, so first you need to know the slope of 5x + y = 6. Subtract 5x from both sides to get slope-intercept form and we see that the slope is -5. That eliminates two of the choices. Because it have to go through (5, 3), plug is 5 for x in each and see which equation gives you 3 for y.

11. It's a demonstration of the Addition Property of Equality. If you add two pairs of congruent segments, the larger segments will be congruent as well.

12. You can't use SSA to prove congruency, except in the special case of Hypotenuse-Leg.

13. The midsegments are half the length of the side they don't touch. The midsegments bisect the sides of the triangle. So there are a whole bunch of congruent segments in that diagram. The perimeter of the trapezoid is 6 + 8 + 5 + 8 + 8 = 35.

14. The exterior angle of a triangle is equal to the sum of the two remote angles. ACR = 125 - 60 = 65.

15. Again! The equation of a circle. The center is (0, -2), so h = 0, and gets omitted, and k = -2, which becomes +2 in the formula. Radius = 5, so 52 = 25.

16. The largest angle of a triangle is across from the longest side. In ABC, if angle B is the largest, AC is the longest side. Choice 1.

17. The more sides that a regular polygon has, the smaller the exterior angles will be because the sum of the exterior angles is always 360.
The fewest sides a polygon can have is 3, a triangle. Divide 360 by 3 and you get 120, which is the largest for regular polygons.
Note: 360 is an impossible answer. You should have crossed it out immediately.

18. They are looking for the lateral area of the cylinder. If it were a soup can, you'd want the area of the label, but not the top and bottom of the can.
The formula is 2*pi*r*h, which is 2*(3.141592...)*9*42, which is 2375.
I don't know why they included 2374 as an answer. If you used 3.1416 for pi, you get 2375.05. If you only used 3.14, you get a smaller answer, which isn't a choice.

19. The shape is a quadrilateral, which has 360 degrees. The angles formed by tangents and radii are right angles (90 degrees each) and you are given 38. Add 90 + 90 + 38 to get 218. Subtract 360-218 to get 142.

20. It a square has a diagonal of 3*(2)^.5, then each side of the square is 3. The perimeter is 12.

21. If you Rotate 90 degrees about the origin and they don't give you a direction then the direction is counter-clockwise!. You move from Quadrant I to Quadrant II, where x is negative and y is positive. The correct answer is (-1, 7).

22. Angles 1 and 2 are supplementary. (They are NOT congruent.) Solve 7x - 36 + 5x + 12 = 180.

12x - 24 = 180, so 12x = 204 and x = 17.

At least this one changes it up a little by making you find the radius, which is 2, so r2 = 4.

24. When a diameter bisects a chord, it is a perpendicular bisector. That means a right angle. That means you can draw an extra radius and make a right triangle and then use the Pythagorean Theorem to find whatever side isn't given.
FE = 17 - 2 = 15. Draw FC, FC = 17. To be "Pythy", CE = 8.

25. Rhombuses (rhombii?) do not have congruent diagonals, unless they are squares. Their diagonals bisect each other.

The center of the circle is (-6, 7), which is in Quadrant II. The radius is 8. That means the circle reaches across the y-axis into Quadrant I and down far enough to cross the x-axis into Quadrant III. Process of elimination, it doesn't go into IV. If you don't believe me, figure that IV is across the origin, and that the origin is (36 + 49)^(.5) = 9.22 away from the center. That's more than 8.

27. No invariant points. That means ALL the points will vary, which means they'll all move.
In other words, no image will have the same co-ordinates (be in the same spot) as the origin.
The three linear reflections all leave a point in the same spot. The point reflection (about the origin) does not.

28. There are four lines which are tangent to both circles at the same time. One touching the "tops", one on the "bottoms" and two more forming an "X" in the middle.

So how did you do?
And what did I get wrong.
Ha! Trick question! I didn't get anything wrong! I just might have made a few inadvertent typos. Yeah! That's the ticket!

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