Question

Pascal wanted the area of the floor to be 54 ft.² in the word still to be 2/3 the lake what was the demons

Answer 1

Answer:

The dimensions of the floor are 9 feet length and 6 feet width.

Step-by-step explanation:

The complete question is

Pascal wanted the area of the floor to be 54 square feet and the width still to be 2/3 the length?what would the dimensions of the floor be.

We know that the area of a rectanlge is $A= w \times l$, where $w$ is width and $l$ is length.

Now, according to the problem, the width is 2/3 of the length, that means

$w=\frac{2}{3}l$

And the area is $A=54 \ ft^{2}$, replacing them, we have

$54=\frac{2}{3}l \times l$

Then, we solve for the length

$54=\frac{2}{3}l^{2}\\ \frac{162}{2}=l^{2} \\ l^{2} =81\\l=\sqrt{81}=9$

So, the length of the floor is 9 feet long.

Now, we use the length value to find the width

$w=\frac{2}{3}(9)=6$

So, the width is 6 feet long.

Therefore, the dimensions of the floor are 9 feet length and 6 feet width.