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 2014
Part II: Each correct answer will receive 2 credits. Partial credit is possible.
29. The coordinates of the endpoints of BC are B(5,1) and C(-3,-2). Under the transformation R90, the image of BC is B'C'. State the coordinates of points B' and C'.
Answer:
The first thing to remember here is if the do not state the direction of the rotation, the default is COUNTER-Clockwise. Don't think of the clock. Think of the Quadrants. Which way do you go to get from I to II? You go counterclockwise.
Second, this is an unusual question because there really isn't work to be shown. You could show formulas, but it isn't necessary. It could be assumed you found it on the graph paper in the back of the book. Basically, each correct endpoint is worth one credit.
The formula for a rotation of R90 ccw is as follows: P(x, y) --> P'(-y, x).
That is, whatever the y coordinate is, change its sign and make it the new x coordinate. Whatever the x coordinate is, make it the y coordinate without changing it. It is better to visualize this and make a sketch than to try and blindly memorize a formula.
B(5, 1) --> B'(-1, 5). Goes from Quadrant I to II.
C(-3, -2) --> C'(2, -3). Goes from Quadrant III to IV.
27. As shown in the diagram below, AS is a diagonal of trapezoid STAR, RA || ST, m∠ATS = 48, m∠RSA = 47, and m∠ARS = 68.
Determine and state the longest side of triangle SAT
Answer:
You don't need to find any lenght, you just need to determine which side, SA, AT, or TS, is the longest side of the triangle.
The longest side of any triangle is the side opposite the greatest angle. In a right triangle, the hypotenuse is always the longest side, and it is across from the right angle, which is the largest angle in a right triangle.
You are given that angle T is 48 degrees. Therefore, SA is NOT the longest side because 48 can be the largest angle. (You know this is because 48 * 3 < 180, right?)
Since RA || ST then RS is a transversal across parallel lines. That means ∠ARS and ∠RST are supplementary angles. Therefore, m∠ARS + m∠RSA + m∠AST = 180
68 + 47 + m∠AST = 180
115 + m∠AST = 180
m∠AST = 65
65 + 48 + m∠SAT = 180
113 + m∠SAT = 180
m∠SAT = 67
The measure of angle SAT is 67, which makes it the biggest angle. Therefore, the side opposite SAT, which is ST, is the longest side.
Be careful with your computations. The numbers are so close together that getting one incorrect number will carry through and affect your final answer. However, if you are consistent, you will only lose a single point for the one error.
Note that guessing ST without showing any work is worth ZERO points even though it is correct. The Regents do not award credit for a guess that is basically a flip of a coin or eeny-meeny-miney.
31. In right triangle ABC shown below, altitude BD is drawn to hypotenuse AC.
If AD = 8 and DC = 10, determine and state the length of AB.
Answer:
The altitude divides the right triangle into two smaller right triangles. What you need to realize is that all three of these right triangles are Similar and therefore the corresponding sides are proportional. Write a proportion and solve it.
AB is the hypotenuse of ABD. AC is the hypotenuse of ABC. AC has a length equal to AD + DC.
In ABD, the leg adjacent angle A is AD. In ABC, the leg adjacent angle A is AB.
So AD / AB = AB / AC.
8 / AB = AB / 18
AB2 = 144
AB = 12
Alternatively, you could have done the following to find the length of altitude BD first. Since ADB is proportional to BDC, we can set up the following proportion:
AD / BD = BD / DC (This is also known as the Right Triangle Altitude Theorem)
8 / BD = DB / 10
BD2 = 80
BD = SQRT(80)
You can now use Pythagorean Theorem to find AB:
82 + SQRT(80)2 = AB2
64 + 80 = AB2
144 = AB2
12 = AB
This second methold is longer but it still a valid and complete answer.
More to come. Comments and questions welcome.
More Regents problems.
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