Question

Given triangle abc with vertices A(2,-1), B(5,6), C(-1,4) as shown. Find the number of square units in the area of triangle abs in simplest form.

Answer 1

Answer:

The area of the triangle is 18 square units.

Step-by-step explanation:

First, we determine the lengths of segments AB, BC and AC by Pythagorean Theorem:

AB

$AB = \sqrt{(5-2)^{2}+[6-(-1)]^{2}}$

$AB \approx 7.616$

BC

$BC = \sqrt{(-1-5)^{2}+(4-6)^{2}}$

$BC \approx 6.325$

AC

$AC = \sqrt{(-1-2)^{2}+[4-(-1)]^{2}}$

$AC \approx 5.831$

Now we determine the area of the triangle by Heron's formula:

$A = \sqrt{s\cdot (s-AB)\cdot (s-BC)\cdot (s-AC)}$ (1)

$s = \frac{AB+BC + AC}{2}$ (2)

Where:

$A$ - Area of the triangle.

$s$ - Semiparameter.

If we know that $AB \approx 7.616$, $BC \approx 6.325$ and $AC \approx 5.831$, then the area of the triangle is:

$s \approx 9.886$

$A = 18$

The area of the triangle is 18 square units.