To solve such a complex shape, its best we split it into simpler shapes:
 ⇒ let's cut along the dashed line to form
   ⇒ a square and a triangle
Area of the triangle
⇒ [tex]\frac{1}{2} *base*height[/tex]
 ⇒ however we don't know the base (dashed line)
  (using Pythagorean theorem) ⇒ [tex]dashed-line=\sqrt{5^2-3^2 }=\sqrt{25-9}=4[/tex]
⇒ now that we know the height, let's solve
  [tex]Area-of-triangle = \frac{1}{2}*4*3=2*3=6ft^2[/tex]  <-- area of triangle
Area of the rectangle
⇒ [tex]length * width[/tex]
- length (longer side of rectangle) --> 9ft
- width (shorter side of rectange) --> dashed line --> 4ft
⇒ so:
 [tex]Area-of-rectangle= 9*4=36ft^2[/tex] <-- area of rectangle)
Combined Area = Area of triangle + Area of rectangle = 6 + 36 = 42ft^2
Answer: 42
Hope that helps!