# What is the area of the triangle formed by the x and y-axes and the line 3x + 6y = 10

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You may use Heron's formula to calculate the area of triangle.

`A = sqrt(p(p-a)(p-b)(p-c))`

You need to determine the lengths of the sides of triangle and the half of the perimeter of triangle.

You should calculate the points of intersection of the line `3x + 6y = 10 ` to x and y axis.

`x=0 =gt 6y=10 =gt y = 5/3 =gt (0, 5/3)`

`y=0 =gt 3x = 10 =gt x = 10/3 =gt (10/3 , 0)`

The vertices of triangle are: `(0,0) , (0, 5/3) , (10/3 , 0)`

You need to use the formula length = `l = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)` to determine the lengths of the sides of triangle:

`a = sqrt((0-0)^2 + (5/3 - 0)^2) =gt a = 5/3`

`b = sqrt((10/3-0)^2 + (0 - 0)^2) =gtb = 10/3`

`c = sqrt((10/3- 0)^2 + (0 - 5/3)^2) =gtc = sqrt (125/9) =gt c = 5sqrt5/3`

Evaluating the half of the perimeter of triangle yields:

`p = (a+b+c)/2 =gt p = (5 + 5sqrt5/3)/2 =gt p =(15 + 5sqrt5)/6`

`A = sqrt(((15 + 5sqrt5)/6)*((5 + 5sqrt5)/6)*((-5 + 5sqrt5)/6)*((15- 5sqrt5)/6))`

You should notice that under the square root there are products that gives differences of squares such that:

`A = sqrt (((15^2 - 125)/36)*((125 - 25)/36))`

`A= (10/6)*(10/6) =gt A = 100/36`

Reducing by 4 yields:

`A = 25/9`

`` **Hence, evaluating the area of triangle by using Heron's formula, yields `A = 25/9` .**

First determine the points at which the line 3x + 6y = 10 intersects the x and y axes. Where the line intersects the x-axis, y = 0

3x = 10 => x = 10/3.

The point of intersection is (10/3, 0)

Where the line intersects the y-axis, x = 0

6y = 10 => y = 5/3

The point of intersection is (0, 5/3)

As the axes are perpendicular to each other, the area can be determined using the formula:

`A = (1/2)*b*h`

=> `A = (1/2)*(10/3)*(5/3)`

=> A = `25/9`

**The required area is 25/9 square units.**