Question 54895
construct a rectangle by dropping perpendiculars from one parallel side to
the other parallel side. Call the height of the perpendiculars h.
Call the other side l, which is the length of the shortest parallel side.
h*l is the area of this rectangle.
Now you have to find lthe areas of the two triangles that are left.
Call the bases of the triangles x and y. Their heights are both h.
Their areas are
(1/2)*x*h and
(1/2)*y*h
Now add the 3 areas
A = h*l + (1/2)*x*h + (1/2)*y*h
Multiply the 1st term by 2/2. This is OK since 2/2 = 1
A = (2/2)*h*l + (1/2)*x*h + (1/2)*y*h
factor out (1/2)*h
A = (1/2)*h*(2l + x + y)
or, what is the same thing,
A = h*((2l + x + y)/2)
This is what is required. (2l + x + y) can be written (l + x + y + l}
l is one side and x + y + l is the other side