Question

the length of the rectaangle is two feet less than 3 times the width.if the area i 65ft^2.find the dimension.

Answer 1

Given that the length of the rectangle is two feet less than 3 times the width and the area is 65 square feet.

To solve this we must make some assumptions.

Let the length of the rectangle be "L" and the breadth be "B".

Therefore,

`l=3b-2`

Also, the area of a rectangle is;

`A=l* b`

Thus,

`\begin{gathered} 65=(3b-2)* b \\ 3b^2-2b=65 \\ 3b^2-2b-65=0 \\ \therefore Break\text{ the expression in to groups} \\ 3b^2+13b-15b-65=0 \\ b(3b+13)-5(3b+13)=0 \\ (b-5)(3b+13)=0 \\ b=5;b=-(13)/(3) \end{gathered}`

Since the breadth can not be a negative number, hence the breadth is 5ft

We would then substitute to get the length.

`\begin{gathered} l=3b-2 \\ l=3(5)-2 \\ l=15-2 \\ l=13 \end{gathered}`

Answer: The dimensions are

`\begin{gathered} \text{Length}=13ft \\ \text{Breadth}=5ft \end{gathered}`