Smekalkin drew a rectangle whose length is three times its width, and the sum of its length and width is 24 cm

Smekalkin drew a rectangle whose length is three times its width, and the sum of its length and width is 24 cm. Find the area of this rectangle.

1) Determine the width of the drawn rectangle if we know that it is 3 times less than its length, and their total length is 24 cm.We form an equation where:

X – width;

3X – length;

24 is the total length of the sides.

X + 3X = 24;

4X = 24;

X = 24/4;

X = 6 cm.

2) Determine the length of the drawn rectangle if we know that it is 3 times longer than its width, and the width is 6 cm.

3 * 6 = 18 cm.

3) Determine the area of the rectangle.

6 * 18 = 108 cm².

Answer: the area of the drawn rectangle was 108 cm².