SOLUTION: Four squares with sides 4, 3, 2 and 1 cm are placed side by side left to right, lower part of the squares are connected forming a base (one line). There is a diagonal line from the

Algebra ->  Rectangles -> SOLUTION: Four squares with sides 4, 3, 2 and 1 cm are placed side by side left to right, lower part of the squares are connected forming a base (one line). There is a diagonal line from the      Log On


   

Question 557109: Four squares with sides 4, 3, 2 and 1 cm are placed side by side left to right, lower part of the squares are connected forming a base (one line). There is a diagonal line from the top left corner of the side 4 cm square to the bottom right of the side 1 cm square. Calculate the area of the trapezoid inside the triangle (diagonal line) of the 3 cm square.
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
 

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



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

Area = %28h%28b%5B1%5D+%2B+b%5B2%5D%29%29%2F2

which with this choice of letters that formula becomes:

Area = %28EF%28DE+%2B+FG%29%29%2F2

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

Area = %283%28DE+%2B+FG%29%29%2F2

So we just need DE and FG

The problem is done with similar triangles.

ᐃDEC ∼ ᐃGFC ∼ ᐃABC

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 ᐃDEC ∼ ᐃABC

%28DE%29%2F%28EC%29 = %28AB%29%2F%28BC%29

%28DE%29%2F6 = 4%2F10

10·DE = 4·6

10·DE = 24

   DE = 24%2F10

   DE = 2.4

Using ᐃGFC ∼ ᐃABC

%28FG%29%2F%28FC%29 = %28AB%29%2F%28BC%29

%28FG%29%2F3 = 4%2F10

10·FG = 4·3

10·FG = 12

   FG = 12%2F10

   FG = 1.2

Substituting in

Area = %283%28DE+%2B+FG%29%29%2F2

Area = %283%282.4+%2B+1.2%29%29%2F2

Area = %283%283.6%29%29%2F2

Area = 5.4 cm²

Edwin