The perimeter of a rectangle is 70 cm. If its length is decreased by 5 cm and its width is increased by 5 cm its area will increase by 50 cm^2. Find the length and width of the original rectangle
Share
The perimeter of a rectangle is 70 cm. If its length is decreased by 5 cm and its width is increased by 5 cm its area will increase by 50 cm^2. Find the length and width of the original rectangle
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Let's say the original length of the rectangle is L and the original width is W. Then, the perimeter is 2L + 2W = 70 cm.
If the length is decreased by 5 cm and the width is increased by 5 cm, the new dimensions of the rectangle become (L - 5) and (W + 5). The increase in the area is 50 cm^2, which can be expressed as:
(L - 5)(W + 5) - LW = 50
Expanding the product on the left-hand side:
LW + 5L + 5W - 5L - 5W - LW = 50
Combining like terms:
5L + 5W = 100
Substituting the value of the perimeter into the equation for 2L + 2W = 70:
2L + 2W = 70
Expanding and solving for L:
L = (70 - 2W)/2
Substituting the value of L into the equation for 5L + 5W = 100:
5((70 - 2W)/2) + 5W = 100
Expanding and solving for W:
35 - W + 5W = 100
6W = 165
W = 27.5
Finally, substituting the value of W back into the equation for L:
L = (70 - 2W)/2
= (70 - 2 * 27.5)/2
= (70 - 55)/2
= (15)/2
= 7.5
So the original dimensions of the rectangle were 7.5 cm (length) and 27.5 cm (width)