Question

The area of a rectangle is 42 ft squared, and the length of the rectangle is 5 ft more than twice the width. Find the dimensions of the rectangle. length and width.​

Answer 1

Answer:

Length = 12 ft

Width = $\frac{7}{2} ft$

Step-by-step explanation:

Given,

Area of rectangle = $42 \: {ft}^{2}$

Width = X

Length = 2x + 5

Now,

$x(2x + 5) = 42$

$2 {x}^{2} + 5x = 42$

$2 {x}^{2} + 5x - 42 = 0$

$2 {x}^{2} + 12x - 7x - 42 = 0$

$2x(x + 6) - 7(x + 6) = 0$

$(2x - 7)(x + 6) = 0$

Either

$2x - 7 = 0$

$2x = 0 + 7$

$2x = 7$

$x = \frac{7}{2}$

Or,

$x + 6 = 0$

$x = 0 - 6$

$x = - 6$

Negative value can't be taken.

So, width = $\frac{7}{2} ft$

Again,

Finding the value of length,

Length = $2x + 5$

$2 \times \frac{7}{2} + 5$

$7 + 5$

$12$

Length = 12 ft

Answer 2

Answer:

length = 12 ft, width = 3.5 ft

Step-by-step explanation:

w = width

l = length = 2w + 5

A = wl = w(2w + 5) = 42

2w² + 5w - 42 = 0

(w + 6)(2w - 7) = 0

w + 6 = 0, w = -6 (dimension cannot be negative)

2w - 7 = 0, w = 3.5

l = 2(3.5) + 5 = 12