Question

A triangle is drawn and squares are created from the lengths of each side. Square A has an area of 19 units 2 and Square B has an area of 45 units 2.
What must be the area of the 3rd square for triangle to have a right angle?
The area of Square C is units 2.

Answer 1

To determine the area of the third square when the triangle has a right angle, we can use the Pythagorean theorem. According to this theorem, in a right triangle with sides \( a \), \( b \), and \( c \) (where \( c \) is the hypotenuse), the following relationship holds:

\[
a^2 + b^2 = c^2
\]

The areas of the squares constructed on the sides of the triangle are equal to the squares of the lengths of those sides. Denote the areas of the squares as follows:

- Area of Square A = \( a^2 = 19 \)
- Area of Square B = \( b^2 = 45 \)
- Area of Square C = \( c^2 \) (unknown)

Given the areas:

\[
a^2 + b^2 = c^2
\]

Substituting the areas:

\[
19 + 45 = c^2
\]

Now calculate the sum:

\[
64 = c^2
\]

To find the area of Square C, we take the square root of both sides:

\[
c = \sqrt{64} = 8
\]

Thus, the area of Square C is:

\[
c^2 = 64 \, \text{units}^2
\]

So the area of the third square (Square C) is \( \boxed{64} \) units².