What is the area of a rectangle if its width is 2/3 of the length and the perimeter is 36 cm?

Express the width (b) of the rectangle in terms of its length (a).
If one quantity is the nth part of another quantity, then this quantity is equal to the product of that quantity and the number n. So the width of the rectangle is:

b = 2/3 * a.

The perimeter of a polygon is equal to the sum of the lengths of all its sides. Since a rectangle is a quadrangle, whose opposite sides are equal and parallel, then its perimeter is:
P = a + b + a + b = 2 * a + 2 * b = 2 * (a + b).

Let’s perform the replacement and substitute the perimeter value given by the condition:

2 * (a + 2/3 * a) = 36.

Let’s solve the resulting equation:

2 * (a + (2 * a) / 3) = 36;

2 * ((3 * a) / 3 + (2 * a) / 3) = 36;

2 * (3 * a + 2 * a) / 3 = 36;

2 * (5 * a) / 3 = 36;

(2 * 5 * a) / 3 = 36;

(10 * a) / 3 = 36;

a = (3 * 36) / (10 * 1);

a = 108/10 cm.

Let’s find the length of the rectangle:
b = 2/3 * a = 2/3 * 108/10 = (2 * 108) / (3 * 10) = 216/30 = 36/5 (cm).

Find the area of ​​the rectangle:
S = a * b = 108/10 * 36/5 = (108 * 36) / (10 * 5) = 3888/50 = 77.76 (cm ^ 2).

Answer: S = 77.76 cm ^ 2.