# find exact area of the triangle ABC?Points: - A (-3,4) - B (2,7) - C (5,1) here's how: - distance AC (base of triangle) - equation of AC - perpendicular-distance from A to BC (this is the...

find exact area of the triangle ABC?

Points:

- A (-3,4)

- B (2,7)

- C (5,1)

here's how:

- distance AC (base of triangle)

- equation of AC

- perpendicular-distance from A to BC (this is the triangle's height)

- area = 1/2 x b x h

### 2 Answers | Add Yours

`c=|AB| = sqrt((-3 - 2)^2+(4 - 7)^2) = sqrt(25+9) = sqrt(34)`

`a=|BC| = sqrt((2 - 5)^2+(7-1)^2) = sqrt(9+36) = sqrt(45)=3sqrt(5)`

`b=|AC| = sqrt((-3 - 5)^2+(4-1)^2)) = sqrt(64+9) = sqrt(73)`

We can use Heron's formula for area

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

#### `s = (sqrt(34)+3sqrt(5)+sqrt(73))/2`

`s-a=(sqrt(34)+3sqrt(5)+sqrt(73))/2 - 3sqrt(5)=(sqrt(34)+sqrt(73)-3sqrt(5))/2`

`s-b=(sqrt(34)+3sqrt(5)-sqrt(73))/2`

`s-c=(3sqrt(5)+sqrt(73)-sqrt(34))/2`

so

`A=sqrt(((sqrt(34)+3sqrt(5)+sqrt(73))/2)((sqrt(34)+sqrt(73)-3sqrt(5))/2)((sqrt(34)+3sqrt(5)-sqrt(73))/2)((3sqrt(5)+sqrt(73)-sqrt(34))/2))``or`

`A=1/4sqrt((sqrt(34)+3sqrt(5)+sqrt(73))(sqrt(34)+sqrt(73)-3sqrt(5))(sqrt(34)+3sqrt(5)-sqrt(73))(3sqrt(5)+sqrt(73)-sqrt(34))`

You can simplify this into

`A=1/4sqrt(2(a^2b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4))`

`A=1/4sqrt(2((45)(73)+(45)(34)+(73)(34))-(45^2+34^2+73^2))`

`A=1/4sqrt(6084)=1/4(78)=39/2=19.5` sq units