Earlier today I set four Area Maze puzzles. Here’s how you solve them.
Area Maze 1
Draw this:
- Line x is (38 + 39)/7 = 77/7 = 11cm
- Line y is 11 - 4 = 7cm
- Line z = 7 + 3 = 10cm
- The missing value is therefore (10 x 6) - 40cm2 = 20cm2
Area Maze 2
Draw this:
- First we can deduce that the length x is 5cm, since 4 x 5 = 20.
- We know that the blue area is therefore 5 x 5 = 25cm.
- So the red rectangle has area 25 + 14 = 39cm.
- Note that 78 is twice 39, so the large rectangle area is double the red rectangle area. Since both rectangles share the same height, the missing value must be double the width of the red rectangle, or 2 x 5 = 10cm
Area Maze 3
Draw this:
- The left rectangle with a blue border covers the rectangle with area 21 and a section of the rectangle below. Imagine placing an identical rectangle to the right of the grey rectangle. The shaded blue area must have area 21, and the dotted blue area must be z.
- The area x is 43 - 21 = 22
- The area of the rectangle with the red border is 4 x 10 = 40. But it is also 29 + y, since a + z = 29. Therefore y = 11
- So x is equal to double y. Which means that any rectangle above the horizontal line must have twice the area of the rectangle beneath the line that it shares a side with.
- The green rectangle must have area double the rectangle underneath it, which has area 29. So the green rectangle has area 2 x 29 = 58
- The missing value is 58 - 21 = 37cm2
Area Maze 4
The answer is 35cm2
But I’ll leave you to prove it!
UPDATE
Here’s a solution to Area Maze 4, posted on twitter by @mathemaniac, aka Australian Numeracy Ambassador Simon Pampena.
I post a puzzle here on a Monday every two weeks. If you like games you might be interested in LOOP, a new type of pool I devised that’s played on an elliptical table. You can also check me out on Twitter, Facebook, Google+ and my personal website.
If you know of any great puzzles that you would like me to set here, get in touch.
