Question

A real estate purveyor purchases a 60{,}00060,00060, comma, 000 square foot \left(\text{ft}^2\right)(ft 2 )(, start text, f, t, end text, squared, )warehouse and decides to turn it into a storage facility. The warehouse's width is exactly \dfrac 2 3 3 2 ​ start fraction, 2, divided by, 3, end fraction of its length. What is the warehouse's width? Round your answer to the nearest foot.

Answer 1

Answer:

200 feet

Step-by-step explanation:

Area of the warehouse $=60,000 ft^2$

Let the length of the warehouse=l

The warehouse's width is exactly  $\dfrac23$ of its length

Therefore: Width of the warehouse$=\dfrac23l$

Area =Length X Width

Therefore:

$\dfrac23l*l=60000\\ Cross multiply\\2l^2=60000*3\\2l^2=180000\\ Divide both sides by 2\\2l^2 \div 2=180000 \div 2\\l^2=90000\\l^2=300^2\\ Length, l=300 feet\\Recall: Width =\dfrac23l\\ Therefore, Width of the warehouse=\dfrac23*300=200 feet$