Answer:
<h2>The dimensions of the floor are 9 feet length and 6 feet width.</h2>
Step-by-step explanation:
The complete question is
<em>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.</em>
<em />
We know that the area of a rectanlge is
, where
is width and
is length.
Now, according to the problem, the width is 2/3 of the length, that means
![w=\frac{2}{3}l](https://tex.z-dn.net/?f=w%3D%5Cfrac%7B2%7D%7B3%7Dl)
And the area is
, replacing them, we have
![54=\frac{2}{3}l \times l](https://tex.z-dn.net/?f=54%3D%5Cfrac%7B2%7D%7B3%7Dl%20%5Ctimes%20l)
Then, we solve for the length
![54=\frac{2}{3}l^{2}\\ \frac{162}{2}=l^{2} \\ l^{2} =81\\l=\sqrt{81}=9](https://tex.z-dn.net/?f=54%3D%5Cfrac%7B2%7D%7B3%7Dl%5E%7B2%7D%5C%5C%20%5Cfrac%7B162%7D%7B2%7D%3Dl%5E%7B2%7D%20%5C%5C%20l%5E%7B2%7D%20%3D81%5C%5Cl%3D%5Csqrt%7B81%7D%3D9)
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](https://tex.z-dn.net/?f=w%3D%5Cfrac%7B2%7D%7B3%7D%289%29%3D6)
So, the width is 6 feet long.
Therefore, the dimensions of the floor are 9 feet length and 6 feet width.