Question

a rectangle has a length that is nine feet less than four times its width. its area is 90 square feet. Algebraically determine the length of its width and length. show the work that leads to your answer.

Answer 1

Let width = 4x

Therefore length must equal 4x - 9

4x + 4x - 9= 90

8x - 9 = 90
(plus 9 to both sides)

8x = 99
(divide both sides by 8)

x= 12.375

width = 4 x 12.375
width = 49.5

length = 49.5 - 9
length = 40.5

Answer 2

Answer:  The length and breadth of the rectangle are 15 feet and 6 feet respectively.

Step-by-step explanation:  Given that a the length of a rectangle is 9 feet less  than four times its width. its area is 90 square feet.

We are to determine the length and breadth of the rectangle algebraically.

Let, 'l' and 'w' represents the length and breadth of the rectangle.

Then, according to the given information, we have

$l=4w-9.$

We know that the area of a rectangle is equal to the product of its length and breadth.

So, the area of the given rectangle will be

$A=l\times w\\\\\Rightarrow 90=(4w-9)\times w\\\\\Rightarrow 90=4w^2-9w\\\\\Rightarrow 4w^2-9w-90=0\\\\\Rightarrow 4w^2-24w+15w-90=0\\\\\Rightarrow 4w(w-6)+15(w-6)=0\\\\\Rightarrow (w-6)(4w+15)=0\\\\\Rightarrow w-6=0,~~~~~4w+15=0\\\\\Rightarrow w=6,~~-\dfrac{15}{4}.$

Since the width of the rectangle cannot be negative, so w = 6 feet.

The length is l = 4w - 9 = 4 × 6 - 9 = 24 - 9 = 15 feet.

Thus, the length and breadth of the rectangle are 15 feet and 6 feet respectively.