The diagonal of the rectangle is 10 cm and its perimeter is 28 cm. Find the sides of the rectangle.

ABCD – rectangle;

AC – diagonal, AC = 10 cm;

PABCD = 28 cm;

Find: AB, AD.

The perimeter of the rectangle PABCD = (AB + AD) * 2, therefore AB + AD = PABCD: 2 = 28: 2 = 14 (cm).

Let AB be equal to x cm, then AD = 14 – x cm.

Consider a triangle ABC. It is rectangular with right angle B.

Hence, by the Pythagorean theorem: AC ^ 2 = AB ^ 2 + BC ^ 2, i.e. 10 ^ 2 = x ^ 2 + (14 – x) ^ 2.

Let’s solve the resulting equation:

10 ^ 2 = x ^ 2 + (14 – x) ^ 2;

x ^ 2 + 196 – 28x + x ^ 2 = 100;

2x ^ 2 – 28x + 196 – 100 = 0;

2x ^ 2 – 28x + 96 = 0;

D = 28 ^ 2 – 4 * 2 * 96 = 784 – 768 = 16; sqrt (D) = 4;

x1,2 = (28 ± 4) / 4;

x1 = 8 (cm), x2 = 6 (cm).

Answer: the sides of the rectangle are 6cm and 8cm.