# Use Heron’s formula to find the area of the triangle joining the points (3, 4), (2, 1) and (6, 0).

### 1 Answer | Add Yours

Heron’s formula gives the area of a triangle with sides a, b and c as Area = `sqrt(s*(s - a)(s - b)(s - c))` where s is the semi-perimeter `s = (a + b + c)/2`

To determine the area of the triangle joining the points (3, 4), (2, 1) and (6, 0) first determine the length of each side. Use the formula for the distance between two points (x1, y1) and (x2, y2), `D = sqrt((x1 - x2)^2 + (y1 - y2)^2)` .

The length of the side between (3, 4) and (2,1) is `sqrt((3- 2)^2 + (4 - 1)^2) = sqrt 10`

The length of the side between (2, 1) and (6, 0) is `sqrt((6-2)^2 + (1 - 0)^2) = sqrt 17`

The length of the side between (3, 4) and (6, 0) is `sqrt((3 - 6)^2 + (4 - 0)^2) = sqrt 25 = 5`

The semi-perimeter is `(sqrt 10 + sqrt 17 + 5)/2`

Area = `sqrt((sqrt 10 + sqrt 17 + 5)/2*(-sqrt 10 + sqrt 17 + 5)/2*(sqrt 10 - sqrt 17 + 5)/2*(sqrt 10 + sqrt 17 - 5)/2)`

=> `(1/4)* sqrt((sqrt 10 + sqrt 17 + 5)*(-sqrt 10 + sqrt 17 + 5)*(sqrt 10 - sqrt 17 + 5)*(sqrt 10 + sqrt 17 - 5))`

=> `(1/4)* sqrt(((sqrt 17 + 5)^2 - 10)(-32 - 5*sqrt 10 + 5*sqrt 17 + 5*sqrt 10 + 5*sqrt 17))`

=> `(1/4)* sqrt((32 + 10*sqrt 17)(-32 + 10*sqrt 17))`

=> `(1/4)* sqrt(100*17 - 32^2)`

=> `(1/4)*sqrt 676`

=> `26/4`

=> 6.5

**The area of the triangle with vertices (3, 4), (2, 1) and (6, 0) is 6.5**

**Sources:**