Given the diagram below, if the area of the shaded region is 103 ft2, what are the dimensions of the inside triangle?

Bigger rectangle:
we can see that
length is 3x-2
so, [tex]L=3x-2[/tex]
width is x+6
so, [tex]W=x+6[/tex]
now, we can find area
[tex]A_b=L\times W[/tex]
[tex]A_b=(3x-2)\times (x+6)[/tex]
Smaller rectangle:
we can see that
length is 2x
so, [tex]L=2x[/tex]
width is x-1
so, [tex]W=x-1[/tex]
now, we can find area
[tex]A_s=L\times W[/tex]
[tex]A_s=(2x)\times (x-1)[/tex]
Area of shaded region:
Area of shaded region = area of bigger rectangle - area of smaller rectangle
[tex]A=A_b-A_s[/tex]
we can plug values
[tex]A=((3x-2)\times (x+6))-((2x)\times (x-1))[/tex]
we are given that area as 103 ft^2
so, we can set it equal
[tex]103=((3x-2)\times (x+6))-((2x)\times (x-1))[/tex]
now, we can solve for x
[tex]103=x^2+18x-12[/tex]
[tex]x^2+18x-12-103=103-103[/tex]
[tex]x^2+18x-115=0[/tex]
now, we can factor it
[tex](x-5)(x+23)=0[/tex]
[tex]x=5,x=-23[/tex]
Since, length can never be negative
so,
[tex]x=5[/tex]
now, we can find dimensions of inner rectangle
Length is
[tex]L=2\times 5[/tex]
[tex]L=10ft[/tex]
width is
[tex]W=5-1[/tex]
[tex]W=4ft[/tex]
so, dimensions are
[tex]L=10ft[/tex]
[tex]W=4ft[/tex]..............Answer