Question

jaye owns a four - sided lot that lies between two parallel streets. if the lot is 44,000 - ft² and has 300 ft frontage on one street and 100 ft frontage on the other, then how far apart are the streets? the streets are feet apart.

Explanation:

Step1: Recall trapezoid - area formula

The area formula for a trapezoid is $A=\frac{(b_1 + b_2)h}{2}$, where $A$ is the area, $b_1$ and $b_2$ are the lengths of the parallel - sides, and $h$ is the height (the distance between the parallel sides). Here, $b_1 = 100$ ft, $b_2 = 300$ ft, and $A=44000$ ft².

Step2: Rearrange the formula to solve for $h$

Starting with $A=\frac{(b_1 + b_2)h}{2}$, we can multiply both sides by 2 to get $2A=(b_1 + b_2)h$. Then, $h=\frac{2A}{b_1 + b_2}$.

Step3: Substitute the given values

Substitute $A = 44000$ ft², $b_1 = 100$ ft, and $b_2 = 300$ ft into the formula $h=\frac{2A}{b_1 + b_2}$. We have $h=\frac{2\times44000}{100 + 300}$.
First, calculate the denominator: $100+300 = 400$. Then, calculate the numerator: $2\times44000=88000$.
So, $h=\frac{88000}{400}=220$ ft.

Answer:

220