In honor of Mathematics Awareness Month, here is another mathematical brain bender from puzzle master Wes Carroll.

**The Unkindest Cut of All, Part 1 of 2**

**Difficulty:** HARD

**Type:** MATH (Spatial)

**Question:**

The area of a square is equal to the square of the length of one side. So, for example, a square with side length 3 has area (3^{2}), or 9. What is the area of a square whose *diagonal* is length 5?

In this puzzle you are working out your spatial visualization (occipital lobes), memory (temporal lobes), and hypothesis generation (frontal lobes).

**Solution**:

12.5

**Explanation**:

I am especially fond of these two ways to solve this problem:

1. Draw the right triangle whose hypotenuse is the square’s diagonal, and whose two legs are two sides of the square. Then use the Pythagorean Theorem (a^2 + b^2 = c^2) to solve for the length of each side. Since two sides are equal, we get (a^2 + a^2 = c^2), or (2(a^2) = c^2) ). Since c is 5, 2(a^2) = 25, making a^2 equal to 25/2, or 12.5. Since the area of the square is a^2, we’re done: it’s 12.5.

2. Tilt the square 45 degrees and draw a square around it such the the corners of the original square just touch the middles of the sides of the new, larger square. The new square has sides each 5 units long (the diagonal of the smaller square), and it therefore has area 25. However, a closer inspection reveals that the area of the larger square must be exactly twice that of the smaller. Therefore the smaller square has area 25/2, or 12.5.

You can now go on to Concentric Shapes: The Unkindest Cut of All, Part 2 of 2

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