Question 557109
<pre>
{{{drawing(800,400, -1,11,-1,5,

rectangle(0,0,4,4),
rectangle(4,0,7,3),rectangle(7,0,9,2),rectangle(9,0,10,1),
green(line(0,4,10,0)), locate(2,0,4), locate(5.5,0,3), locate(8,0,2), locate(9.5,0,1),locate(-.2,2,4) 




 )}}} 

Let's chop off the tops and use some lettering:

{{{drawing(800,400, -1,11,-1,5,
locate(4,0,E), locate(7,0,F), locate(4,2.7,D),locate(7,1.5,G),
locate(0,4.3,A), locate(0,0,B), locate(10,0,C),locate(9,0,H),
triangle(0,0,0,4,0,0), triangle(0,0,10,0,0,0),
triangle(4,0,4,2.4,4,0),triangle(7,0,7,1.2,7,0),triangle(9,0,9,.4,9,0),
green(line(0,4,10,0)), locate(2,0,4), locate(5.5,0,3), locate(8,0,2), locate(9.5,0,1),locate(-.2,2,4) )}}}

We want to find the area of trapezoid DEFG.
The formula for the area of a trapezoid is

Area = {{{(h(b[1] + b[2]))/2}}}

which with this choice of letters that formula becomes:

Area = {{{(EF(DE + FG))/2}}}

and we are given that EF = 3, so the formula further becomes

Area = {{{(3(DE + FG))/2}}}

So we just need DE and FG

The problem is done with similar triangles.

&#5123;DEC &#8764; &#5123;GFC &#8764; &#5123;ABC

AB = 4 (given)
BC = BE+EF+FH+HC = 4+3+2+1 = 10, 
EC = EF+FH+HC = 3+2+1 = 6, 
FC = FH+HC = 2+1 = 3

Using &#5123;DEC &#8764; &#5123;ABC

{{{(DE)/(EC)}}} = {{{(AB)/(BC)}}}

{{{(DE)/6}}} = {{{4/10}}}

10·DE = 4·6

10·DE = 24

   DE = {{{24/10}}}

   DE = 2.4

Using &#5123;GFC &#8764; &#5123;ABC

{{{(FG)/(FC)}}} = {{{(AB)/(BC)}}}

{{{(FG)/3}}} = {{{4/10}}}

10·FG = 4·3

10·FG = 12

   FG = {{{12/10}}}

   FG = 1.2

Substituting in

Area = {{{(3(DE + FG))/2}}}

Area = {{{(3(2.4 + 1.2))/2}}}

Area = {{{(3(3.6))/2}}}

Area = 5.4 cm²

Edwin</pre>