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, January 2014
Part II: Each correct answer will receive 2 credits. Partial credit is possible.
29. The diameter of a sphere is 5 inches. Determine and state the surface area of the sphere, to the nearest hundredth of a square inch.
Answer:
The formula for the Surface Area of a Sphere is in the back of the booklet: SA = 4πr2. The radius is HALF the diameter. To the nearest hundredth of a square inch means TWO decimal places.
SA = 4πr2 = 4 (3.141592...)(2.5)2 = 78.539...
The Surface Area is 78.54 square inches.
Note that if you didn't use enough accuracy for pi, such as only using 3.14, you will make a mistake.
30. Using a compass and straightedge, construct the perpendicular bisector of AB.
[Leave all construction marks.]
Answer:
To construct a perpendicular bisector, start at point A. Open you compass more than halfway to B. Make an arc centered on A. Move the compass to B without changing the size. Make an arc that overlaps the first arc. Use the straightedge to draw the perpendicular bisector.
31. The endpoints of AB are A(3,-4) and B(7,2). Determine and state the length of AB in simplest
radical form.
Answer:
Use the Distance Formula, or Pythagorean Theorem. (They're related.) You can even make a sketch to find the distances between the x-coordinates and the y-coordinates.
d = SQRT ( (7 - 3)2 + (-4 - 2)2 )
d = SQRT ( 42 + (-6)2 )
d = SQRT ( 16 + 36 )
d = SQRT ( 52 )
You are asked for "Simplest Radical Form" so DO NOT give a decimal, not one rounded to two places, not one with 8 significant digits followed by three dots.
The prime factors of 52 = 2 * 2 * 13 = 22 * 13
SQRT (22 * 13) = 2 * SQRT (13)
More to come. Comments and questions welcome.
More Regents problems.
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