Question

The area of a rectangle is 42ft^2, and the length of the rectangle is 11 less than three times the width. Find the dimensions of the rectangle.

Answer 1

Answer:

7 ft ×6 ft

Step-by-step explanation:

Area of rectangle= length ×width

Let the length and width of the rectangle be L and W ft respectively.

Form 2 equations using the given information:

LW= 42 -----(1)

L= 3W -11 -----(2)

Substitute (2) into (1):

(3W -11)(W)= 42

Expand:

3W² -11W= 42

3W² -11W -42= 0

$\boxed{x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} }$

Applying quadratic formula:

$W= \frac{ - ( - 11)± \sqrt{( - 11) {}^{2} - 4(3)( - 42) } }{2(3)}$

$W= \frac{ 11± \sqrt{625 } }{6}$

$W= \frac{ 11± 25}{6}$

W= 6 or W= $- \frac{7}{3}$ (reject)

Substitute W= 6 into (1):

L(6)= 42

L= 42 ÷6

L= 7

Thus the dimensions of the rectangle is 7 ft ×6 ft.