The length of the rectangle is 5 cm longer than its width. On two adjacent sides of this rectangle

The length of the rectangle is 5 cm longer than its width. On two adjacent sides of this rectangle, squares are built, the difference in area of which is 85 cm3. Find the area of the rectangle.

1. The vertices of the rectangle A, B, C, D. The square PADK adjoins the side AD, the square MNBA adjoins the side AB.

2. We take the length of the AB side as x. Side length AD – (x + 5).

3. The area of the MNBA square is x².

4. The area of the square PADK is equal to (x + 5) ².

5. Taking into account that the difference of the squares’ areas is 85 cm², we draw up the equation:

(x + 5) ² – x² = 85;

x² + 10x + 25 – x² = 85;

10x = 60;

x = 6 cm.

AB = 6 cm.

AD = 6 + 5 = 11 cm.

6. The area of the rectangle ABCD = AB x AD = 6 x 11 = 66 cm ².