School-Safe Puzzle Games

## Semi-Circle Geometry Puzzle

What is the area of white (without the semi-circle moon shadows)?

Each square is one square inch.

How do you prove it? (pi=3.142)

Created for Smart-Kit by Forrest Frantz

For the teachers following, this definitely makes for a good printable puzzle.

If you can figure out the answer, feel free to enter it into the comment section below. Submissions will remain hidden for a day or 2 so as to not spoil the fun!

### 32 Comments to “Semi-Circle Geometry Puzzle”

1. Ari | Profile

The area between x=f(y)+c and x=f(y) is /f(y)+cy – /f(y) = cy

Here we have two pairs of functions with offsets of 1 and 2 between them (c=1, c=2)
So the larger shaded area is A = cy = 1*6 and the smaller is 2*2
The white area is 4*6 – 1*6 – 2*2 = 24 – 10 = 14in^2

2. lennondogg | Profile

24-[3.5*(pi)+4] = 9.003

3. foger1979 | Profile

This is my best lunch break guess
Area = 14

Taking the big white cirlce and figuring its area minus the shaded part
Big white cirlce
A = 3.142(3×3)/2 = 3.142(9)/2 = 28.278/2 = 14.139
A = 3.142(1×1)/2 = 3.142/2 = 1.571
plus 4 squares = 5.571
smaller white circle
3.142(1×1)/2 = 3.142/2 = 1.571
All together
14.139 – 5.571 + 1.571
Area of large white circle = 10.139

Now take the larger shaded area
A = 3.142(3×3)/2 and add 6 for the area of sqaures on the right side
3.142(9)/2 + 6
28.278/2 + 6 = 14.139 + 6 = 20.139

Total area of rectangle is 24

For the white space:
24 – 20.139 + 10.139

= 14

4. suineg | PUZZLE MASTER | Profile

Ok, I think it can be solved this way:
Total Area= 4*6= 24 square inch
Green Area= Green Ellipse – White Circle + Green Square
White Area= Total Area – Green Area
Lets calculate Green Area:
Green Ellipse = 3.142*3*4/2–>18.852
White Circle= 3.142*9/2—->14.139
Green Square= 2*2—>4
Green Area= 18.852-14.139+4—>8.713
White Area= 24-8.713= 15.287
This is my approach, but I know that someone can find a more exact way to calculate that area, because it can be said that the half of green ellipse its incomplete in the frame of the total area, but I dont want to go that deeper.

5. mroth01 | Profile

Area of white = total area – green square in center-right – (1/2) * (area of large ellipse – area of small ellipse)

area of ellipse = pi*a*b ; a = 4 for large ellipse, 3 for small ellipse; b = 3

Area of white = 24 – 4 – (1/2) *(12pi – 9pi)

= 20 – (3*pi/2) = 15.287 sq in

6. lograh | Profile

I’ll guess 14 square inches.
To start, I made the assumption that each curve was an arc of a perfect circle (without that, solving this becomes much harder). I noticed that the blue hemisphere in boxes (2,3) and (2,4) is the same area as the white one in (4,3) and (4,4) so they can be swapped to leave a nice square blue region in the middle of the right side. I also noticed that the leftmost curve was a match for the curve immediately to its right, just shifted one unit left (if it were really a concentric arc with the smaller, the radial distance between them would be constant). So you can shift the white areas to the far left right by one unit and you are left with a nice contiguous white zone of 18 square inches. Subtract the four blue that are left at the middle of the right side and you are left with 14 square inches of white.

Or did I miss something, since I never needed to use pi (either exact or your rounding)?

7. jasc | Profile

14

6*3 (shifting white circle left eliminates blue sickle)less 4 (interchanging blue half circle w/ white makes 2*2)

8. cmetzger4 | Profile

Area = 22-pi

or

18.859

9. falwan | Profile

A= {[16(3.142)-9(3.142)]/2}+4
= 3.5(3.142)+4
= 11.957+4
= 15.957

10. falwan | Profile

Of course the area of the white portion of the shape is:

A(W)= 24 – 15.957 = 8.043

11. doar823 | Profile

Well…one way of doing this I think is by elimination…if the “moons” are perfectly shaped then on the 1st, 2nd, 5th, and 6th rows, there is one square worth of green in each. meaning 3 square inches of white->12square inches…in the middle two rows, there is 3 full squares of green and 1square inch of white.
This means the area of white is 14 square inches. Being a science major I try not to use pi as much as possible

12. falwan | Profile

Another approach:

area of large shaded portion=6 sq. in.
area of small shaded portion=4 sq. in.

Area of white portion = 24 – 10 = 14 sq. in.

Would say this approach is more convinsing because of irregularity of outer arc.

Thank You.

13. jay | Profile

I’m sorry disregard my last comment. I meant 14 sq. inches…

14. da Moose | Profile

Area of small blue crescent:
2 x 3 – ((3.142 x 1^2)/2) = 4.429
4.429 – 2 + ((3.142 x 1^2)/2) = 4 sq in

White area inside large blue crescent:
((3.142 x 3^2)/2) – 4 = 10.139 sq in

White area outsite large blue crescent:
3 x 6 – ((3.142 x 3^2)/2) = 3.861 sq in

Therefore, total white area is:
10.139 sq in + 3.861 sq in = 14 sq in

15. T | Profile

I’m really bad at maths so I’m sure I’m way off…

Area of Hypothetical LARGE GREY semicircle in (0,0) and (3,6) = a1

a1 = 9pi/2
= 14.139

Area of LARGE GREY HALFMOON in (0,0) and (4,6) is a1 + area of rectangle (3,0) and (4,6) minus the area of white half circle = a2

a2 = (a1 + 6) – 9pi/2
= 14.139 + 6 – 14.139
= 6

Smaller grey area small half circle, plus two squares plus two squares minus small half circle = a3

a3 = pi/2 + 2 + (2-pi/2)
= 1.571 + 2 + (2-1.571)
= 4

Total white area is rectangle in (0,0) and (4,6) minue (a2 + a3)

Total white Area = 4×6 – 4 + 6
= 24 – 10
= 14

Total area without semi-circle moon shadows (or total white area) is 14 square inches.

16. sharhan | Profile

Outer big semi-circle – Inner big semi-circle=6 sequares

small shadow = 4 sequaers (obvious)

White Areas = 24 – 10 = 14

No need for pi !!

17. vikram.mankal | Profile

14 sq. inches.
The greenparts form 10 out of the total 24 sq. inches.

Name the squares as:
rows – a,b,c,d,e and f
columns – 1,2,3 and 4

The green of C2 can be superimposed on the white of C4 to complete the square.
Similarily, D2 on D4.

Also, the remaining smaller green part of C2 and D2 can be superimposed on C1 and D1 respectively to complete those squares. This way a total of 10 greeen squares can be formed. Therefore leaving a total of 14 white squares.

18. rathersane | Profile

My reasoning can be found on an index card I wrote it all down on, scanned, and posted here:

19. Hazlewood | Profile

24-((((Area of big circle/2)+6)-(Area of big circle/2))+((Area of small circle/2)+2+(2-(Area of small circle/2)))=16

Sorry, no time to draw this out. Cool little puzzle that looks more tricky than it really is. The grid reminded me of the triangle puzzle that is disassembled, rearranged, and there is a mysterious piece missing. Drove me nuts for a while.
These types of puzzles are a blast though. Thanks.

20. suineg | PUZZLE MASTER | Profile

Another approach more intuitive and less mathematically exact(no need of pi value) than my first post is this:
In the grid:

First and second row: if you overlap the square in a single square you would make aproximately 1 green square per row

Third and fourth row: same logic but a little more complex overlap the superior part of the second square with the first and the inferior part with the fourth
3 green square per row= 6 squares

Fifth and sixth row: the same that in the first 2 rows
2 green squares
Total green= 10 squares–> 10 square inches

Total area= 24 square inches–> White Area= 14 square inches

Again Mathematically, because I cannot see my previous post it was like this:
1)Green Half Ellipse= pi*A*B/2–> 3.142*6= 18.852
2)Half Green circle= pi*(square r) /2= 3.142*1/2= 1.571–> Half little White circle
3)Half White circle= pi*9/2= 14.139
4)Green Big Square (IFComplete)=4
5)Green Area=(1)-(3)+(4)-(2)+(2)=18.852-14.139 + 4= 8,713
Total Area= 24
White Area = Total Area- Green Area
White Area= 24-8.713= 15.287

The first was not that bad for a visual aproximation.

21. ian pedder | Profile

White area = 14

Total area is 24 and blue area is 10 leaving 14 for white.

The small blue semi circle can be moved right to the white semi cicle making a blue square of 4 units.

The blue crescent shape has an area of 6. This can be considered made up of a big semi circle plus the 6 square vertical strip on right hand side less the white semi circle which cancels out.

Hence blue area = 10

22. darkfaith_452 | Profile

since all the green squares equate to 10 full squares, whats left ova are 14 white squares out of the 24.

since a sqaure is 1 inch^2 in area then 14 by 1 square inch is 14 square inch. am i right….it seems to sinmple or maybe i didnt get the question.

23. pope | Profile

14 square inches.
First row has one full block shaded out
Second row has one full block shaded out (2 total)
Third row has 3 full blocks shaded out. (5 total)
The bottom three rows mirror the top three rows (10 total)
There’s 10 full blocks shaded out, which is equal to 10 sq. inches. The entire area of the diagram is 24 sq inches (6×4 = 24). 24-10 = 14 sq. inches.

24. RK | Profile

I thought this one was very very tricky….

The answer is 14 square inches

25. RK | Profile

as several of you noted, you can solve it visually by moving areas to complete whole squares

26. RK | Profile

27. Tommy | Profile

Dumb way:
(9pi/2+6)(leftmost big semicircle plus 6 below)-9pi/2(other biggie)+4(the rest)=10
24-10=14

Good Way
The principle is pretty much you can subtract a figure from anywhere on the figure and the remainder is always the same. So, I subtracted the big white semicircle from the big blue semicircle and the six squares to the left of it. This gives me 6. Do the same with the smaller ones and get 4. Voila, the answer is 24-10=14.

The way that RK showed requires more brainpower to do(and I’m lazy so I take shortcuts) but may be slightly easier to see.

28. jbyers2 | Profile

“Sliding out the tab” reveals the area of the small green semi-circle and adjacent area is 4 units; a similar approach reveals the area of large semi-circle is 6 units, making a total of 10 units.

29. scriptar | Profile

Part of the problem with this question is that each of the “squares” that make up the grid are not actual squares and so the picture is misleading. The purpose of the squares is to show you how many units wide and long the shapes are. You can’t actually take the shaded parts of the area between the outer ellipse and the circle and piece them together to make 6 squares although you can make a square with the inner half-circle. Plus, there are the black lines grid that take up area within the 4 inch wide rectangle. To solve this problem, you must make some assumptions about the shapes and use the formulas for the areas of a rectangle, an ellipse (a half-ellipse in this case), a circle (again, half a circle), and a square. I will assume that the “area of white” includes everything except the shaded areas, where parts of the shade are under the lines and parts of the white are under the lines.

area of white = area of rectangle – shaded area

shaded area = (area of half-ellipse – area of half-circle) + (area of square made from wedge of half-circle cut from square and translated to left of square)

area of half-ellipse = (pi * A * B) / 2

area of half-circle = (pi * r * r) / 2

area of square = l * l

area of rectangle = 4 * 6 = 24 square inches

shaded area = (pi * A * B / 2) – (pi * r * r / 2) + (l * l)

shaded area = pi / 2 * (A * B – r * r) + l * l

shaded area = pi / 2 * (4 * 3 – 3 * 3) + 2 * 2

shaded area = pi / 2 * (12 – 9) + 4

shaded area = pi * 3 / 2 + 4

shaded area = pi * 1.5 + 4

If pi = 3.142, then shaded area = 8.713 square inches

Area of white = 24 – 8.713 = 15.287 square inches

30. scriptar | Profile

I realized that the outer shape is not an ellipse when I couldn’t easily duplicate the curve in Illustrator so my above answer goes out the window. So instead, I decided to find the exact answer.

First, I resized the image to 4 inches by 6 inches, converted it to black and white, and removed the lines. Here’s the black-and-white-only image: http://scriptar.com/images/bw_geometry.gif

Next, I wrote a little script that counts the number of white pixels. Here is my output:
Width: 384
Height: 576
Total Pixels: 221184
White Pixels: 123473
White to Total Ratio: 0.55823658130787
Total area: 24 square inches
Total white area: 13.3976779513889 square inches

So that settles it. Let them eat Pi.

31. RK | Profile

sorry that appeared twice, still a couple of bugs in the comment system