-1. The area of EOF is a quarter of a circle with point O at the center of the circle. The line segment AC is 8 inches long and the line segment BC is 6 inches long.

How many square inches are in the shaded areas? (round to the nearest tenth)



-2. Assume the radius of each circle is 5 inches. How many square inches are in the shaded areas? (round to the nearest hundredth)



-3. If we paint each face of a 4 foot cube with the color red, then cut the cube into 64 one-foot cubes....

How many small cubes will have no red color on any side of the cube?
How many small cubes have 1 face with red color?
How many small cubes have 3 faces with red color?

3. 8 will have no red face at all
24 will have one red face
8 will have 3 red faces

(I think...!)

#1

r = radius = sqrt(6^2 + 8^2) = 10

a = area quarter circle = pi * r^2 / 4 = 78.5398...

shaded area = a - 6*8 = 30.5398... = 30.5 sq.in.

Perfect solves, guys!

2. 228.3 sq.in. Explanation may be a bit tedious. Object is to find the area of 1 half of a shaded part.Consider one circle.Join the four points on the circumference to form a square. Each side of the square is 5sqrt2. Area of the square is 50. The area of one circle is 25pi. Subtracting these and dividing by 4 gives the area of half of one of the shaded parts. (25pi-50)/4. There are 32 such shaded halves.

Area of shaded parts = 32(25pi-50)/4 = 200pi-400 = 228.3 sq.in.

Spot on, manxie!