Question

Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.

Answer 1

The dimensions of the rectangle are either 8 meters by 5 meters or 5 meters by 8 meters, based on the given perimeter and area.

Step-by-step explanation:

To find the dimensions of a rectangle given the perimeter and the area, we need to set up a system of equations based on the formulas for perimeter and area of a rectangle. The perimeter (P) of a rectangle is given by the formula P = 2l + 2w, where l is the length and w is the width. The area (A) is given by the formula A = l × w.

In this question, the perimeter is 26 meters and the area is 40 square meters, so we have the following equations:

1. 2l + 2w = 26 (perimeter equation)
2. lw = 40 (area equation)

First, we can simplify the perimeter equation by dividing everything by 2, getting l + w = 13.
Next, we can express w in terms of l by rearranging the simplified perimeter equation: w = 13 - l.

Now, substitute w = 13 - l into the area equation lw = 40:

1. l(13 - l) = 40
2. 13l - l^2 = 40
3. l^2 - 13l + 40 = 0

This is a quadratic equation which can be factored to (l - 8)(l - 5) = 0, giving us two possible lengths: l = 8 meters or l = 5 meters. If l = 8 meters, then w = 5 meters, and if l = 5 meters, then w = 8 meters.

Therefore, the dimensions of the rectangle can be 8 meters by 5 meters or 5 meters by 8 meters.

Answer 2

`x,y-\ dimensions\ of\ rectangle\\\\\ Perimeter\\\\ P=2x+2y\\ P=26\\\\ 26=2x+2y\ |:2\\ 13=x+y\\ x=13-y\\\\\ Area:\\\\ A=x*y\\ A=40\\ 40=x*y\\\\ Substituting\ x=13-y\\\\ 40=(13-y)*y\\ 40=13y-y^2\\y^2-13y+40=0\\ \Delta=13^2-4*1*40=169-160=9\\ √(\Delta)=\sqrt9=3\\\\ y=(13-3)/(2)=(10)/(2)=5\ \ or y=(13+3)/(2)=(16)/(2)=8\\x=13-y\\x=13-5=8\ or\ x=13-8=5\\\\Dimensions\ are\ 5m\ and\ 8m.`