Question

A triangle is drawn and squares are created from the lengths of each side. Square A has an area of 16 units2
and Square B has an area of 80 units2
. What must be the area of the 3rd square for triangle to have a right angle? A standard proof image of the Pythagorean theorem. (1 point

Answer 1

To determine the area of the third square in order for the triangle to have a right angle, we can utilize the Pythagorean theorem. The theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.

Let:
- \( A \) be the area of the first square, which has area \( 16 \) units². Thus, the side length \( a \) of Square A is \( \sqrt{16} = 4 \) units.
- \( B \) be the area of the second square, which has area \( 80 \) units². Thus, the side length \( b \) of Square B is \( \sqrt{80} = 4\sqrt{5} \) units.

Let \( C \) be the area of the third square, and the side length \( c = \sqrt{C} \).

According to the Pythagorean theorem:
\[
c^2 = a^2 + b^2
\]
Substituting the values:
\[
c^2 = 16 + 80 = 96
\]
Thus, the area of square C must be:
\[
C = c^2 = 96 \text{ units}^2
\]

Therefore, the area of the third square must be **96 units²** for the triangle to have a right angle.