Question

A triangle is drawn and square are created from the length of each side. Square A has an area of 18 units ^2 and Square B has an area of 90 units ^2. What must be the area of the 3rd square for triangle to have a right angle?

The area of Square C = ? Units ^2

Answer 1

To determine the area of the third square (Square C) such that the triangle formed by the sides of the squares is a right triangle, we can use the Pythagorean theorem.

According to the theorem, for a triangle with sides of lengths \(a\), \(b\), and \(c\), where \(c\) is the length of the hypotenuse, the following relationship holds:
\[
a^2 + b^2 = c^2
\]

In this case, we have:

- The area of Square A is \(18\) units\(^2\), hence the side length (let's denote it \(a\)) is:
\[
a = \sqrt{18} = 3\sqrt{2}
\]

- The area of Square B is \(90\) units\(^2\), hence the side length (let's denote it \(b\)) is:
\[
b = \sqrt{90} = 3\sqrt{10}
\]

Let's denote the area of Square C as \(c^2\). According to Pythagorean theorem, we need to find \(c^2\) such that:
\[
a^2 + b^2 = c^2
\]

Calculating \(a^2\) and \(b^2\):
\[
a^2 = 18
\]
\[
b^2 = 90
\]

Now, summing \(a^2\) and \(b^2\):
\[
a^2 + b^2 = 18 + 90 = 108
\]

Therefore, for the triangle to be a right triangle, the area of Square C must be:
\[
c^2 = 108
\]

Thus, the area of Square C is:
\[
\boxed{108} \text{ units }^2
\]