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.
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.
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