**6x+5y+14=0...........(1)**

** 6y=5x-29................(2)**

**11y+x=18....****............(3)**

**We will find intersection points foe (1) and (2).**

**==> 6x+5y+ 14 = 0**

**==> y= (5x-29)/6 **

**==> 6x + 5(5x-29)/6 +14 = 0**

==> 6x + (25x -145)/6 +14 = 0

==> (36x +25x -145)/6 = -14

==> 61x -145 = -84

==> 61x = 61

==> x= 1 ==> y= (5x-29)/6 = (5-29)/6= -4

==> Let the intersection point be A(1,-4).

Now we will find intersection between (1) and (3).

==> 6x+5y+14= 0

==> x= 18-11y

==> 6(18-11y) + 5y +14 = 0

==> 108 - 66y +5y +14 = 0

==> -61y = -122

==> y= 2 ==> x = 18-11y = 18-22 = -4

Then let the intersection point between (1) and (3) be B(-4,2).

Now we will find the intersection point between (2) and (3)

==> 5x-6y -29= 0

==> x= 18-11y

==> 5(18-11y) - 6y -29 = 0

==> 90 - 55y -6y -29= 0

==> -61y +61= 0

==> y= 1 ==> x = 18-11y= 18-11 = 7

Then, the intersection between (2) and (3) is C(7,1)

Then, we have the triangle ABC such that:

A(1,-4), B(-4,2), and C(7,1)

We will find the length of the sides.

==> `AB= sqrt((1+4)^2 + (-4-2)^2)= sqrt(25+36)= sqrt(61)`

`==> BC= sqrt((-4-7)^2 +(2-1)^2) = sqrt(121+1)= sqrt122`

`==> AC= sqrt((1-7)^2 +(-4-1)^2)= sqrt(36+25)=sqrt122`

We notice that the triangle is an isosceles triangle .

==> height =` sqrt((sqrt122)^2 - (sqrt61/2)^2)`

`==> Height = sqrt(122- 61/4) = sqrt(427/4)= sqrt427 / 2`

`==> A= (1/2)*base*height`

`==> A = (1/2)* sqrt61 * sqrt427/2`

`==> A = sqrt(61*427)/4 = sqrt(26047) / 4 = 40.35`

`==> A= 40.35`

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