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.

Question

asked Feb 5, 2022 14.8k views

1 Answer

Jli · Answer 1 · 2022-02-10T10:22:33+0000

Answer:

The area of the triangle is 18 square units.

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:

$A = √(s\cdot (s-AB)\cdot (s-BC)\cdot (s-AC))$ (1)

s = (AB+BC + AC)/(2) (2)

Where:

- Area of the triangle.

- 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.