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

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(s-a)(s-b)(s-c))` where `s=(a+b+c)/2`

No using the formula to find the distance between two points
`A(x_A,y_A) , B(x_B,y_B)`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(3,2)` `B(5,10)` , `C(8,4)`

After that, we substitute to Heron formula.

2nd Solution

We know that if `(x_1,y_1), (x_2,y_2)` and `(x_3,y_3)` are the vertices of the triangle, then the area of the triangle is given by:

Area of the triangle`= (1/2) |{(x2-x1)(y2+y1) +(x3-x2)(y3+y1)+(x1-x3)(y1+y2)}|`

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

Area of the triangle`= (1/2) |{(5-3)(10+2) +(8-5)(4+2)+(3-8)(2+10)}|=abs(1/2(24+18-60))=9`

Method 1: Geometric

`triangle ABC = PQRS - (triangleAPB+triangleBQC+ACRS)`

`PQRS = 5xx10 = 50`
`triangle APB = 1/2(8xx2) = 8`
`triangle BQC = 1/2(3xx6) =9`
`ACRS = (2+4)/2xx5 =15`

`triangle ABC = 50 -(8+9+15) = 50 -32 = 18`

Method 2: Herons Formula
Using the Pythagorean Theorem we can calculate the lengths of the sides of `triangle ABC`
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