Question: how do we work the problem on Chapter 14 test #4.
Answer from Harold Jacobs:
Since all 16 white regions are identical, we can find the area of one of them and multiply it by 16.
This figure represents one of the 16 small squares and one of the white regions.
Since the diameter of the large circle in the figure shown on the test page is 4 units, the side of the square of one of the small squares is 1 unit.
We can find the area of region B by finding the area of regions B & C combined (a quarter of a circle) and subtracting the area of region C (a triangle):
> The area of B & C combined is1/4 (pi r squared) or 1/4 (pi 1 squared) = 1/4 pi
> The area of C is1/2 r squared or 1/2 1 squared = 1/2So B = (B + C) – C = 1/4 pi – 1/2 and the total area of the 16 white regions is16(1/4 pi – 1/2) = 4 pi – 8.
Question from Kerry:
The answer key for the Geometry Midterm Review #74 says the answer is square root of 80, but shouldn’t it be the square root of 52? It says (4-0) squared + (8-2)squared = 16+64, but it should be = 16+36, shouldn’t it?
Answer from Dr. Callahan:
You are correct. The original Enhanced Teacher’s Guide for Geometry (ISBN: 978-0-7167-5607-1) had the correct answer. When it was republished as Answers to Exercises for Geometry (ISBN: 978-1-61999-116-3) they changed the answer.
The correct answer is SQRT(52).