Imara used these steps to find the length of the hypotenuse of the right triangle. step 1: find the area of the square with side lengths of 20: 400
step 2: find the area of the square with side lengths of 15: 225
step 3: find the sum of the areas of the two squares: 625
step 4: state the length of the hypotenuse: 625
which best describes imara’s error?

solution:

as the sides of right triangle that is either base = 15 cm or 20 cm, altitude = 20 cm or 15 cm

so, we have constructed square on the sides having length 15 cm and side having length 20 cm.

area of square having side length 20 cm is (20 cm)²= 400 cm²

area of square having side length 15 cm is   (15 cm)² = 225 cm²

sum of areas of two squares = 400 + 225 = 625

so, length of hypotenuse = √625 = 25 cm

fourth statement is incorrect that describes imara's error.

the length of the hypotenuse: ≠ 625. it should be cm

the two may have that, even a breakdown is probably offered