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