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Math Help - Diagonals of an isosceles trapezium

  1. #1
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    Diagonals of an isosceles trapezium

    There is an isosceles trapezium with both the oblique sides equal to 8 cm, and the parallel sides being 4, and 9 cm respectively.

    It is required to find the diagonals. Now, my answer is 10 cm. Can anyone please verify this ?
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  2. #2
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    Hello Basal
    Quote Originally Posted by Basal View Post
    There is an isosceles trapezium with both the oblique sides equal to 8 cm, and the parallel sides being 4, and 9 cm respectively.

    It is required to find the diagonals. Now, my answer is 10 cm. Can anyone please verify this ?
    Not quite. If the height of the trapezium is h cm, then
    h^2 = 8^2+2.5^2\;*
    and if the diagonal is of length d cm, then
    d^2 = h^2 + 6.5^2
    =8^2+2.5^2 + 6.5^2
    \Rightarrow d = \sqrt{112.5}
    =10.61 cm (2 d.p.)
    Grandad

    * See correction, below!
    Last edited by Grandad; April 20th 2010 at 08:01 AM.
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  3. #3
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    Quote Originally Posted by Basal View Post
    There is an isosceles trapezium with both the oblique sides equal to 8 cm, and the parallel sides being 4, and 9 cm respectively.

    It is required to find the diagonals. Now, my answer is 10 cm. Can anyone please verify this ?
    This is correct.

    Note to Grandad: 8cm is not the height, it is the sidelength (or maybe you're just making a mistake I don't understand)
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  4. #4
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    Hello Laurent
    Quote Originally Posted by Laurent View Post
    This is correct.

    Note to Grandad: 8cm is not the height, it is the sidelength (or maybe you're just making a mistake I don't understand)
    You're right - the answer is 10 cm, although I don't understand what you mean here. I didn't say the height was 8 cm. I had a sign wrong. I should have said
    h^2 = 8^2 - 2.5^2
    of course!

    Grandad
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  5. #5
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    Thanks for verification. I had to be sure because for some reason the textbook gives the answer as sqrt(73), which is clearly wrong.

    Side question : How do you do the root thingy ?
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