Please show all work!

-1. The diameter of the big circle is 40 inches. What is the total circumferences of all 5 circles? Assume the line segment passes through the center of all of the circles. (Pi = 3.14)



-2. The curve is composed by connecting 6 quarter circumferences with different sizes as shown in the figure. Assume the length of one side of the small square in the middle is 1 inch, what is the length of the curve? (Pi = 3.14)




-3. A 24cm horizontal line is drawn at a tangent to the inside circle, as shown. Calculate the shaded area.



-4. Area ABC is 40 square feet. The length of line segment BD is 1/4 of line segment AB. The length of line segment EC is 1/3 of line segment AC. How many square feet is the shaded area CDE?

2. Circumference of a circle = 2*pi(r)
Circumference of a quarter circle is 2*pi(r)/4 = pi(r)/2
quarter circle 1 has a radius of 1 inch and each successive quarter circle has a radius increase of 1 inch (by adding 1 side of the square each time)
The sum of all 6 quarter circle circumferences is:
pi(1)/2 + pi(2)/2 + pi(3)/ +pi(4)/2 +pi(5)/2 +pi(6)/2 = pi(21)/2
32.97 inches

Pandora, I've missed your Math Fun posts

That's a perfect solve, manxman! I'll try to post math puzzles more often.

Numero 3:

u = inner circle radius
v = outer circle radius

a = area inner circle = pi(u^2)
b = area outer circle = pi(v^2)

shaded area = b - a

Dropping a perpendicular from the chord line
to the center (both circles have same center)
creates a right triangle with legs = u and 12,
and the hypotenuse = v.

So v^2 = u^2 + 12^2 = u^2 + 144 [1]

shaded area = pi(v^2) - pi(u^2)
= pi(v^2 - u^2)
= pi(u^2 + 144 - u^2) : substituting [1]
= pi(144)
= 3.14(144)
= 452.16

Btw, there is an "area of annulus" theorem
that states the results of above:
area annulus = pi * (chord/2)^2

But you stated "show all work"

1. it will be the same as that for the big circle. Since the formula is pi x Diameter, the total diameters of the small circles is the same as the diameter of the big circle.

Well done, Denis. Thank you for showing all work.

DocPaul, you're absolutely correct! So what is the answer?

4.
Area of BCD = (Area of ABC)/4 = 40/4 = 10 square feet
Area of ACD = (Area of ABC)-(Area of BCD) = 40-10 = 30 square feet
Area of CDE = (Area of ACD)/3 = 30/3 = 10 square feet

Perfectly done!

you still want the answer? LOL 125.663704 thereabouts.

That needs to be doubled, Doc:
"all the circles" includes the big one.

Represented using algebra:
let diameters of any 3 inner circles = a, b, c;
then 4th inner circle's diameter = 40 - a - b - c

total circumference of the inner circles
= pi(a) + pi(b) + pi(c) + pi(40 - a - b - c)
= pi(a + b + c + 40 - a - b - c)
= pi(40)
= same as outer circle

And that's a wrap! Well done, guys!