The length of the rectangle is 6 cm longer than its width. After the length was increased by 9 cm and the width by 12 cm

| January 13, 2021 | education

The length of the rectangle is 6 cm longer than its width. After the length was increased by 9 cm and the width by 12 cm, the area of the rectangle increased 3 times. Find the perimeter of the original rectangle.

1. Let’s denote the length of the rectangle by X, and the width by Y.
2. Then, by the condition of the problem, X = Y + 6, and the area S1 = X * Y = Y ^ 2 + 6 * Y.
3. The area of ​​the enlarged rectangle S2 = (X + 9) * (Y + 12) = (Y + 15) * (Y + 12) = Y ^ 2 + 27 * Y + 180.
4. By the condition of the problem S2 = 3 * S1. From this equality we obtain a quadratic equation:
2 * Y ^ 2 – 9 * Y – 180 = 0.
5. The discriminant of the equation D ^ 2 = 81 + 1440 = 1521. D = 39.
6. The roots of the equation Y = – 30/4 and Y = 12. The negative root does not correspond to the condition of the problem, therefore the width of the rectangle is 12 cm, and the length is 12 + 6 = 18 cm.
7. The perimeter of the rectangle is P = 2 * (X + Y) = 2 * (12 + 18) = 60 cm.
Answer: The perimeter of the original rectangle is 60 cm.



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