Well this shape is a combination of 2 triangles, so we'll find the area of both of them and add them together.
[tex]A_{triangle} = \frac{1}{2}bh[/tex] where [tex]b[/tex] is the base of the triangle and [tex]h[/tex] is the height.
For the triangle on the right, the base is 8 and the height is 8.
[tex]A_{triangle} = \frac{1}{2}*8*8 = 32[/tex]
So the area of the rightmost triangle is [tex]32yds^{2}[/tex]
For the one on the right, we know the height is 8 as it's the same height as the other one. It shows that [tex]\frac{1}{2}[/tex] of the base is 7, so the whole base is 14. So we plug it into the equation:
[tex]A_{triangle} = \frac{1}{2}*14*8 = 58[/tex]
So the area of the second triangle is [tex]56yds^{2}[/tex]
So to find the area of the whole shape, we need to add the two together:
[tex]56yds^{2} + 32yds^{2} = 88yds^{2}[/tex]
So the area of the shape is [tex]88yds^{2}[/tex]
I'm assuming that you hadn't learned the area of a trapezoid, but that also could have been used.
