How do you find the dimensions of a rectangle whose perimeter is 52 m and whose area is 160 m?

Question

Maverick Smith · Accepted Answer

The dimension of rectangle is #l=16 , b=10# Perimeter of rectangle is #p=2(l+b) =52#, #l# is length and #b# is breadth. Area of rectangle is #A=l*b =160 :. l=160/b ; :. 2(160/b+b)=52 or (160/b+b)=26 or (160+b^2)/b=26 or 160+b^2=26b or b^2-26b+160=0 or b^2-16b-10b+160=0 or b(b-16) -10(b-16) =0 or (b-16)(b-10)=0 :. b=16 or b=10# If #b=16 ; l=160/16=10 # and if #b=10 ; l=160/10=16 # The dimension of rectangle is #l=16 , b=10# [Ans]

Avery Allen · Answer

undefined Let $ l $ be the length and $ w $ be the width of the rectangle.

Given:
Perimeter $ P = 2l + 2w = 52 $
Area $ A = lw = 160 $

From the perimeter equation:
$$ 2l + 2w = 52 $$
$$ l + w = 26 $$
$$ l = 26 - w $$

Substitute $ l $ from the perimeter equation into the area equation:
$$ (26 - w)w = 160 $$
$$ 26w - w^2 = 160 $$
$$ w^2 - 26w + 160 = 0 $$

Using quadratic formula:
$$ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
where $ a = 1 $, $ b = -26 $, and $ c = 160 $
$$ w = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(1)(160)}}{2(1)} $$
$$ w = \frac{26 \pm \sqrt{676 - 640}}{2} $$
$$ w = \frac{26 \pm \sqrt{36}}{2} $$
$$ w = \frac{26 \pm 6}{2} $$

The possible values for $ w $ are:
$$ w_1 = \frac{26 + 6}{2} = 16 $$
$$ w_2 = \frac{26 - 6}{2} = 10 $$

When $ w = 16 $, then $ l = 26 - 16 = 10 $
When $ w = 10 $, then $ l = 26 - 10 = 16 $

Thus, the dimensions of the rectangle are either $ 16 \, \text{m} \times 10 \, \text{m} $ or $ 10 \, \text{m} \times 16 \, \text{m} $.