Thread: Solving for side of square given diagonal

Can someone correct me?

Problem: The diagonal of a square quilt is 4 times the square root of 2. What is the area of the quilt in square feet?

The answer is 4. My book suggests that we use the property of a 45 45 90 triangle to solve this. I understand that method. However, I initially attempted to solve it by using Pythagorean theorem with the diagonal being the hypotenuse. I came up with 4 x 2^(1/2) = 2 x (a^2) I proceeded to divide the left side by 2, and I did not come up with 4 for the answer. Why is this?

If you want to use Phytagoras then, let be the side of the square quilt therefore:

Solve this equation for .

Originally Posted by benny92000

Can someone correct me?

Problem: The diagonal of a square quilt is 4 times the square root of 2. What is the area of the quilt in square feet?

The answer is 4. <--- That is the side-length and not the area
My book suggests that we use the property of a 45 45 90 triangle to solve this. I understand that method. However, I initially attempted to solve it by using Pythagorean theorem with the diagonal being the hypotenuse. I came up with 4 x 2^(1/2) = 2 x (a^2) I proceeded to divide the left side by 2, and I did not come up with 4 for the answer. Why is this?

1. Draw a sketch.

2. The quilt is placed inside a square of the side-length d.

3. You can prove (by congruent triangles) that the area of the quilt is



4. Plug in the value for d and you'll get the given result.

5. If you want to calculate the side-length then

Siron has calculated the sides of the square when what was asked for was the area, s^2.

Originally Posted by cathectio

Siron has calculated the sides of the square when what was asked for was the area, s^2.

no kidding ...

Siron most probably assumed the OP was capable of determining the area of a square once he/she determined the side length.

Effective tutoring sometimes requires providing just enough information to let the OP solve the problem (or screw it up) on their own.