A triangle is drawn and squares are created from the lengths of each side. Square A has an area of 25 units² and Square C has an area of 70 units². What must be the area of the 3rd square for triangle to have a right angle?

Answer 1

To determine the area of the third square, we need to find the length of the corresponding side of the triangle.

Let's denote the length of the sides of the triangle as a, b, and c, with a being the length of the side corresponding to Square A, b being the length corresponding to the unknown square, and c being the length corresponding to Square C.

According to the problem, the area of Square A is 25 units^2. Therefore, the side length a can be found by taking the square root of 25:

a = √25 = 5 units

Similarly, the area of Square C is 70 units^2. Therefore, the side length c can be found by taking the square root of 70:

c = √70

Now, in a right-angled triangle, the squares of the lengths of the two shorter sides added together must equal the square of the length of the longest side. This is known as the Pythagorean theorem:

a^2 + b^2 = c^2

Substituting the known values, we have:

(5^2) + b^2 = (√70)^2

25 + b^2 = 70

b^2 = 70 - 25 = 45

b = √45 = √(9 x 5) = 3√5 units

Therefore, the length of the third side of the triangle (corresponding to the unknown square) is 3√5 units. To find the area of the third square, we square this length:

Area of the third square = (3√5)^2 = 3^2 × (√5)^2 = 9 × 5 = 45 units^2

So, the area of the third square must be 45 units^2 for the triangle to have a right angle.