Chapter 8.4, Problem 35AYU

Cost of a Triangular Lot The dimensions of a triangular lot are 100 feet by 50 feet by 75 feet. If the price of such land is $\$3$ per square foot, how much does the lot cost?

To determine

To find: How much does the lot cost?

Answer to Problem 35AYU

Solution:

$\text{Cos}t = \$5446.38$

Explanation of Solution

Given:

The dimensions of a triangular lot are 100 feet by 50 feet by 75 feet. The price of such land is $\$3$ per square foot.

Formula used:

Area of the triangle $= \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{a + b + c}{2}$

Calculation:

The dimensions of a triangular lot are 100 feet by 50 feet by 75 feet.

Let $a = 100 feet, b = 50 feet and c = 75 feet$

$s = \frac{a + b + c}{2}$

$S = \frac{100 + 50 + 75}{2}$

$S = \frac{225}{2}$

$Area = \sqrt{\left(\frac{225}{2}\right) \left(\frac{225}{2} - 100\right) \left(\frac{225}{2} - 50\right) \left(\frac{225}{2} - 75\right)}$

$Area = 1815.46$ ${\text{ft}}^2$

$\text{Cos}t = 1815.46 ft² * \$3$

$\text{Cos}t = \$5446.38$