Around a right triangle are 3 suqares rhe square on the left C is 9m^2, the square on the bottom D is 16m^2, what is the area of a the square on the other side and the length of b the hypotenuse of the triangle?
To find the area of the square on the other side, we can use the Pythagorean theorem to determine the length of the hypotenuse of the triangle.
Since the square on the left (C) is 9m^2 and the square on the bottom (D) is 16m^2, we can find the side lengths of the squares by taking the square root:
Side length of square C = sqrt(9) = 3m
Side length of square D = sqrt(16) = 4m
Let's denote the side length of the square on the other side as x, and the length of the hypotenuse as b.
Since the squares are arranged around a right triangle, the side lengths of the squares will also be the lengths of the sides of the triangle.
Now, using the Pythagorean theorem, we have:
a^2 + b^2 = c^2
(3 + 4)^2 + b^2 = x^2
7^2 + b^2 = x^2
49 + b^2 = x^2
Since the side length of the square on the other side is also equal to x, the area of the square will be x^2.
We know that the area of square C + square D + square on the other side = total area of the squares around the triangle
9 + 16 + x^2 = 25 + x^2
25 + x^2 = 25 + b^2
x^2 = b^2
Therefore, the area of the square on the other side is equal to the square of the hypotenuse of the triangle.
So, the area of the square on the other side is 25m^2, and the length of the hypotenuse b of the triangle is 5m.