Chapter 11.1, Problem 55AYU

To determine

To find: The dimensions of a rectangular floor if the length is twice the width and the perimeter of the floor is 90 feet.

Answer to Problem 55AYU

$\text{Length }=30\text{ feet};\text{ Width }=15\text{ feet}$.

Explanation of Solution

Given:

Length is twice the width of the rectangular floor.

$\text{Perimeter }=90\text{ feet}$.

Calculation:

Let the width of the rectangular floor be $x$ feet.

Then the length of the floor is $2x$ feet.

Perimeter of the floor $=90\text{ feet}$.

We know that in a rectangle,

$\text{Perimeter }=2\times \left(\text{length }+\text{ width}\right)$

Thus we get,

$2\times \left(2x+x\right)=90$

$2\times 3x=90$

$6x=90$

$x=\frac{90}{6}$

$x=15$

Therefore the width of the rectangular floor is 15 feet.

Then the length of the rectangular floor becomes $=2\times 15=30$ feet.

Hence the dimensions, that is, the length and width of the rectangular floor are 30 feet and 15 feet respectively.