Question

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:

A----G-------------D
|.........|..................|
|.........|..................|
E----F..................|
|.........|..................|
|.........|..................|
|.........|..................|
|.........|..................|
B----H-------------C

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 ?

Answer 1

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.

Answer 2

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 ?

Answer 3

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

Answer 4

Thank you for all your help.

Answer 5

O_O