Area reduction problem

• Nov 21st 2005, 10:23 AM
ronheid
Area reduction problem
I have a rectangle which consists of a width of 10", and a length of 40". THis gives me a total area of 400" (w x l= a). If I reduce the area by a percentage, lets say 10% my reduced area is now 360" (400 - 40 = 360.) What is the formula to determine the reduced size of the width and the length.

W = 10"
L = 40"
A = 400

reduced rectangle 10% of area
A2 = 360
W2 = ?
L2 = ?

• Nov 21st 2005, 11:45 AM
hpe
Quote:

Originally Posted by ronheid
I have a rectangle which consists of a width of 10", and a length of 40". THis gives me a total area of 400" (w x l= a). If I reduce the area by a percentage, lets say 10% my reduced area is now 360" (400 - 40 = 360.) What is the formula to determine the reduced size of the width and the length.

W = 10"
L = 40"
A = 400

reduced rectangle 10% of area
A2 = 360
W2 = ?
L2 = ?

Reduce both width and length by 5.13% (leaving 94.87%). Since $\displaystyle 0.9487 \cdot 0.9487 \approx 0.90$, this will reduce the area by 10%.
The general formula: To reduce the area by x%, reduce the linear dimensions by $\displaystyle 100\left(1 - \sqrt{1 - \frac{x}{100}}\right)\%$.