Question

How do you find the area of a triangle whose vertices are (2,3),(5,7),(4,-2)?

Answer 1

Area `= 11.5` (square units)

Explanation:

This could be done in many different ways.

Method 1: Using Pythagorean and Heron's Formulas
You could use the Pythagorean theorem to calculate the distance between each pair of points (i.e. the lengths of the sides of the triangle)
then use Heron's formula `A=sqrt(s(s-a)(s-b)(s-c))` [where `s` is the semi-perimeter and `a,b, c` are the length of the sides].

Method 2: Geometric Method

The area of the `triangleABC` is equal to to
the area of `square ABED -(triangleACD + triangleBEC)`

`squareABED = (5+9)/2*3 = 21`

`triangleACD = (5*2)/2 = 5`

`triangleBEC= (9*1)/2 = 4.5`

`triangleABC = 21 - (5+4.5) = 11.5`

`color(white)("XXX")`[Read `square` as "trapezoid"; I couldn't find a better symbol]