Question

A city planning commission recently voted to restrict the size of home remodels by limiting the floor area to lot area ratio to maximum of 0.60 to 1. Under these guidelines,

A) what would be the maximum allowable size of a remodel on an 11,400 sq ft lot?

B) what size lot would be required in order to create a 5040 sq ft remodel?

Answer 1

Answer:

Step-by-step explanation:

From the given information:

The ratio of the limiting floor area to lot area is 0.60 to 1

i.e.

= 0.60 : 1

For instance, let's take that the remodel size as p sq.ft on 1140 sq.ft

Then, the ratio = $\dfrac{p}{11400}$

The proportion of these equations is as follows:

$\dfrac{p}{11400} = \dfrac{0.60}{1}$

$p \times 1 = 11400 \times 0.60$

$p = 11400 \times \dfrac{60}{100}$

$p =\dfrac{ 11400 \times 60}{100}$

$p =\dfrac{ 114 \times 100 \times 60}{100}$

$p = 114 \times 60$

p = 6840 sq ft

Thus, the maximum allowable size of a remodeled house is 6840 sq.ft

b.

Recall that, the size of the home remodeled by limiting floor to lot area is

0.60:1

Then the proportion equation form is as follows:

$\dfrac{0.60}{1}= \dfrac{5040}{x}$

By cross multiplying

$0.60 \times x = 5040 \times 1$

$x = \dfrac{5040 \times 1}{0.60 }$

$x = \dfrac{5040 \times 100}{60 }$

$x = \dfrac{84 \times 60 \times 100}{60 }$

$x =84 \times 100$

x = 8400 sq.ft

Hence, the size of lot area is 8,400 sq.ft