Question

1. Find the length of a rectangular lot with a perimeter of 124 feet if the length is two more than twice the width.

2. Find the length of a rectangular lot with a perimeter of 50 feet if the length is five more than the width.

Answer 1

ANSWER TO QUESTION 1


Let the width be
$w$
units.



Twice the width will be

$2w$

Two more than twice the width will be

$2w + 2$



We were told that the length is 2 more than twice the width. This means that,

$l = 2w + 2$


This implies that,

$w = \frac{l - 2}{2} - - eqn(1)$


The formula for finding the perimeter of a rectangle is given by:


$perimeter = 2w + 2l - - eqn(2)$




We substitute 124 for the perimeter and equation (1) into equation (2)



$124 = 2( \frac{l - 2}{2} ) + 2l$


This implies that,

$124 = l - 2 + 2l$

$\Rightarrow 124 + 2 = l + 2l$



$\Rightarrow 126 = 3l$




We divide both sides by 3 to obtain,


$l = 42$

Therefore the length is 42 feet.





ANSWER TO QUESTION 2


Let the width be
$w$


units.


Five more than the width will be


$w + 5$


We were told that the length is 5 more than the width. This means that,


$l = w + 5$


We make w the subject to obtain,


$w = l - 5 - - eqn(1)$



The perimeter is given by,

$perimeter = 2w + 2l - - eqn(2)$

We substitute 50 for the perimeter and equation (1) into equation (2)


$50 = 2(l - 5) + 2l$

Dividing through by 2 gives,


$25 = l - 5 + l$



$\Rightarrow 25 + 5 = l + l$



$\Rightarrow 30 = 2l$


We divide both sides by 2 to get,

$l = 15$


Therefore the length is 15 feet.