Gr. 9 Math (continued)
posted by olav .
I don't see a problem but I will rewrite the structure.
Make two rectangles attached together.
Call the first rectangle ABHG (AG top, BH bottom).
Call the second rectangle GHCD (HC top, GD bottom).
Draw a horizontal line from line AB to line GH and name the points E on line AB and F on line GH.
example:
AGD
...........................
...........................
EF..................
...........................
...........................
...........................
...........................
BHC
Rectangles AEFG, EBHF and GHCD are similar.
The length of AE is 4cm.
The length of EB is 9cm.
What is the area of rectangle GHCD ?
What are the lengths of HC and EF ?

Given AE=4, EB=9,
If AE≠EB, the only way that AEFG and EBHF are similar is that AE<EF<EB, which means really that
AEFG is similar to FEBH to ensure the vertices correspond.
If that's the case,
Let EF=x, then
AE/x = x/EB, or
x² = AE*EB = 36
EF = x = 6 (negative root rejected)
Now for the rectangle GHCD, there are two ways to be similar to rectangle AEFG, i.e.
HC=(2/3)GH, or
HC=(3/2)GH
I will let you complete the problem. 
I looked at it another way.
I used rectangle FEBH; area = 9 x EF.
I used rectangle AEFG; area = 4 x EF.
Ratio of area FEBH to AEFG = 2.25
The square root of area is ratio of the perimeters = 1.5
Perimeter of AEFG = 2 x 9 + 2 x EF
Perimeter of FEBH = 2 x 4 + 2 x EF.
Since FEBH is 1.5 larger than AEFG then
1.5 x (8 + 2EF) = 18 + 2EF,
12 + 3EF = 18 + 2EF,
1EF = 6.
Now I know ratio of AE / EF = 2/3.
It works out the same FE / HF = 2/3.
Therefore GH / HC = 2/3,
13 x 3 / 2 = HC = 19.5.
Therefore the area of GHCD =
13 x 19.5 = 253.5
Is this correct ? 
That is correct, but HC=19.5 is only one of the two possible solutions.
If HC=13*(2/3)=26/3, the rectangles are still similar (but turned 90°).
So the other solution is
Area = 13*(26/3) = 112.67 
Thank you for all your help.

O_O