# Question 94454

Nov 8, 2017

$\therefore$ length$= 25 f t .$ and width $= 12 f t .$

#### Explanation:

Let the width of the rectangle be $w$ ft.
So, according to given data:

length $l = 1 f t . + 2 w$

Area of rectangle $A = l \times w = 300 f t {.}^{2}$

$\therefore$ 300 Ft^2 = w xx l

$\implies w \times \left(2 w + 1\right) = 300$

$\implies 2 {w}^{2} + w = 300$

$\implies 2 {w}^{2} + w - 300 = 0$

We need two such numbers which add upto to give coefficient of middle term (+1) and whose product is equal to the product of coefficient of first term and last term( 2 x 300 =600).

Two such numbers are $25 \mathmr{and} - 24$

$\implies 2 {w}^{2} - 24 w + 25 w - 300 = 0$

$\implies 2 w \left(w - 12\right) + 25 \left(w - 12\right) = 0$

$= . \left(2 w + 25\right) \left(w - 12\right) = 0$

$\therefore \left(2 w + 25\right) = 0 \mathmr{and} \left(w - 12\right) = 0$

$\implies 2 w = - 25 \implies w = - \frac{25}{2} = - 12.5$

But width cannot be negative. So, we take the other value,

$\implies w = 12$

So, the dimensions are :

$w = 12$ft.
and
$l = 1 + 2 w = 1 + 24 = 25$

$\therefore$le$n$gth$= 25 \mathmr{and} w i \mathrm{dt} h = 12$