Chapter 8.4, Problem 11AYU

To determine

To find: The area of the given triangle.

Answer to Problem 11AYU

Solution:

$Area \approx 9.56$

Explanation of Solution

Given:

Formula used:

$K = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

Calculation:

$a = 9, b = 6, c = 4$

Area of the triangle is given by

$K = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

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

$s = \frac{9 + 6 + 4}{2} = \frac{19}{2}$

$K = \sqrt{\left(\frac{19}{2}\right) \left(\frac{19}{2} - 9\right) \left(\frac{19}{2} - 6\right) \left(\frac{19}{2} - 4\right)}$

$K = \frac{\sqrt{1463}}{4}$

$K \approx 9.56$

$Area \approx 9.56$