The diagonal of the rectangle is 13 and its width is 7 less than its length. Find the sides of the rectangle.

Let the length of the rectangle be equal to X cm, then, by condition, its width will be equal to (X – 7) cm.

In a right-angled triangle ACD, by the Pythagorean theorem, we express the hypotenuse AC.

AC ^ 2 = AD ^ 2 + CD ^ 2.

13 ^ 2 = X ^ 2 + (X – 7) ^ 2.

169 = X ^ 2 + X ^ 2 – 14 * X + 49.

2 * X ^ 2 – 14 * X – 120 = 0.

X ^ 2 – 7 * X – 60 = 0.

Let’s solve the quadratic equation.

D = b ^ 2 – 4 * a * c = (-7) 2 – 4 * 1 * (-60) = 49 + 240 = 289.

X1 = (7 – √289) * (2/1) = (7 – 17) / 2 = -10 / 2 = -5. (Doesn’t fit because <0).

X2 = (7 + √289) * (2/1) = (7 + 17) / 2 = 24/2 = 12.

AD = BC = 12 cm, then AB = SD = 12 – 7 = 5 cm.

Answer: The sides of the rectangle are 12 cm and 5 cm.