Gr. 9 Math (continued)

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.



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 ?

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asked by olav
  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

    I will let you complete the problem.

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  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 ?

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    posted by olav
  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

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  4. Thank you for all your help.

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    posted by olav
  5. O_O

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    posted by Alex

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