What is the area of a triangle whose vertices are the points with coordinates (3,2) (5,10) and (8,4)?

Refer to explanation

Explanation:

1st solution

We can use Heron formula which states

The area of a triangle with sides a,b,c is equal to

$S = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where $s = \frac{a + b + c}{2}$

No using the formula to find the distance between two points
$A \left({x}_{A} , {y}_{A}\right) , B \left({x}_{B} , {y}_{B}\right)$which is

(AB)=sqrt((x_A-x_B)^2+(y_A-y_B)^2

we can calculate the length of sides between the three points given
let say $A \left(3 , 2\right)$ $B \left(5 , 10\right)$ , $C \left(8 , 4\right)$

After that, we substitute to Heron formula.

2nd Solution

We know that if $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$ are the vertices of the triangle, then the area of the triangle is given by:

Area of the triangle$= \left(\frac{1}{2}\right) | \left\{\left(x 2 - x 1\right) \left(y 2 + y 1\right) + \left(x 3 - x 2\right) \left(y 3 + y 1\right) + \left(x 1 - x 3\right) \left(y 1 + y 2\right)\right\} |$

Therefore the area of the triangle whose vertices are $\left(3 , 2\right) , \left(5 , 10\right) , \left(8 , 4\right)$ is given by:

Area of the triangle$= \left(\frac{1}{2}\right) | \left\{\left(5 - 3\right) \left(10 + 2\right) + \left(8 - 5\right) \left(4 + 2\right) + \left(3 - 8\right) \left(2 + 10\right)\right\} | = \left\mid \frac{1}{2} \left(24 + 18 - 60\right) \right\mid = 9$

Oct 1, 2015

$18$

Explanation:

Method 1: Geometric

$\triangle A B C = P Q R S - \left(\triangle A P B + \triangle B Q C + A C R S\right)$

$P Q R S = 5 \times 10 = 50$
$\triangle A P B = \frac{1}{2} \left(8 \times 2\right) = 8$
$\triangle B Q C = \frac{1}{2} \left(3 \times 6\right) = 9$
$A C R S = \frac{2 + 4}{2} \times 5 = 15$

$\triangle A B C = 50 - \left(8 + 9 + 15\right) = 50 - 32 = 18$

Method 2: Herons Formula
Using the Pythagorean Theorem we can calculate the lengths of the sides of $\triangle A B C$
then we can use Heron's Formula for the area of a triangle given the lengths of its sides.

Because of the number of calculations involved (and the need to evaluate square roots), I did this in a spreadsheet:

Again (fortunately) I got an answer of $18$ for the area