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Math Help - Quadrilaterals

  1. #1
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    Quadrilaterals

    Hello everybody, this is my first post!

    How do I find the radius of a circle that is in side a diamond/ rombus? The quadrilateral has 2 diagonals of 10 and 20cm. I have got as far as finding that if you cross the diagonals you get 4 triangles and i have found that the sides of the diamond are 22.36cm. I cant get any further! Help pleaase.

    Thank you.
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  2. #2
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    Assuming I am interpreting your problem correctly, I get half what you got for the length of a side: 5\sqrt{5}\approx{11.18}

    Because, \sqrt{10^{2}+5^{2}}=5\sqrt{5}\approx{11.18}

    This means we can let ED=x and then AE=5\sqrt{5}-x

    CD=10 and AB=20. Therefore, OC=OD=5 and OA=OB=10.

    Then, the radius of the circle is OE.

    And, AD=AC=CB=BD=5\sqrt{5}

    OE=\sqrt{5^{2}-x^{2}}=\sqrt{10^{2}-(5\sqrt{5}-x)^{2}}

    Solving for x we see x=\sqrt{5}

    Check my figures. Perhaps too much New Year's cheer.

    If I misinterpreted the diagonal lengths, adjust accordingly.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  3. #3
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    Using galactusí diagram, note that OE is the mean proportional between DE and DA.
    So OE^2  = 5\sqrt 5 x - x^2 ; moreover OE^2  + x^2  = 25.
    This gives OE = \sqrt {20}
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  4. #4
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Sweeties View Post
    i have found that the sides of the diamond are 22.36cm.
    The side is actually 25 cm. The sum of the squares of the diagonals of a rhombus is equal to four times the square of the length of a side.

    Also the ratio of the shorter diagonal to the longer diagonal is equal to \tan{\frac{\theta}{2}}, where \theta is an acute interior angle of the rhombus. In this case, \tan{\frac{\theta}{2}}=\frac{1}{2}; hence \sin{\frac{\theta}{2}}=\frac{1}{\sqrt{5}}. According to the diagram, the radius you want is 10\sin{\frac{\theta}{2}}=\frac{10}{\sqrt{5}}=2\sqr  t{5}\,\mathrm{cm}.
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  5. #5
    Senior Member JaneBennet's Avatar
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    Rhombus fact: If the length of each side of a rhombus is a and \theta is an acute interior angle, the shorter diagonal has length 2a\sin{\frac{\theta}{2}} and the longer diagonal has length 2a\cos{\frac{\theta}{2}}.

    Hence: (i) In a rhombus with sides of fixed length, the sum of the squares of the lengths of the diagonals is constant. (ii) In a rhombus with fixed interior angles, the ratio of the lengths of the diagonals is constant.
    Last edited by JaneBennet; December 31st 2007 at 06:58 PM.
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  6. #6
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    Quote Originally Posted by Sweeties View Post
    Hello everybody, this is my first post!

    How do I find the radius of a circle that is in side a diamond/ rombus? The quadrilateral has 2 diagonals of 10 and 20cm. I have got as far as finding that if you cross the diagonals you get 4 triangles and i have found that the sides of the diamond are 22.36cm. I cant get any further! Help pleaase.

    Thank you.
    Hello,

    only an additional remark: You are dealing with 4 congruent right triangles and you are looking for the length of the height in one of those triangles. The height of one triangle is the radius of the inscribed circle.

    If you have a right triangle with the legs a and b and the hypotenuse c then the height h is calculated by:

    h = \frac{a \cdot b}{c}

    With your problem:

    \begin{array}{l}a = 5\\b = 10\\c = 5\sqrt{5}\end{array}

    Therefore h = r = \frac{5 \cdot 10}{5\sqrt{5}} = 2 \cdot \sqrt{5}
    Attached Thumbnails Attached Thumbnails Quadrilaterals-inkrsrad_rhomb.gif  
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  7. #7
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    Quote Originally Posted by earboth View Post
    Hello,

    only an additional remark: You are dealing with 4 congruent right triangles and you are looking for the length of the height in one of those triangles. The height of one triangle is the radius of the inscribed circle.

    If you have a right triangle with the legs a and b and the hypotenuse c then the height h is calculated by:

    h = \frac{a \cdot b}{c}

    With your problem:

    \begin{array}{l}a = 5\\b = 10\\c = 5\sqrt{5}\end{array}

    Therefore h = r = \frac{5 \cdot 10}{5\sqrt{5}} = 2 \cdot \sqrt{5}
    h = \frac{a \cdot b}{c}

    I haven't seen that formula before, it seems the easiest way of getting the answer, i was trying to use the formula for area of a triangle to find something like this but it didn't work.
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  8. #8
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    Quote Originally Posted by Sweeties View Post
    h = \frac{a \cdot b}{c}

    I haven't seen that formula before, it seems the easiest way of getting the answer, i was trying to use the formula for area of a triangle to find something like this but it didn't work.
    Prove the bigger right angle triangle is similar to the inner right angle triangle. Then use similarity property, that sides are proportional.

    Happy New Year
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  9. #9
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    Quote Originally Posted by Sweeties View Post
    h = \frac{a \cdot b}{c}

    I haven't seen that formula before, it seems the easiest way of getting the answer, i was trying to use the formula for area of a triangle to find something like this but it didn't work.
    Hello,

    this formula only works with right triangles.

    The area of a right triangle can be calculated by:

    a = \frac12 \cdot c \cdot h ........ [1] or

    a = \frac12 \cdot a \cdot b ........ [2]

    and therefore: \frac12 \cdot c \cdot h = \frac12 \cdot a \cdot b........... Solve this equation for h and you'll get:

    \boxed{  h = \frac{a \cdot b}{c}}
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  10. #10
    Lord of certain Rings
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    Quote Originally Posted by earboth View Post
    Hello,

    this formula only works with right triangles.

    The area of a right triangle can be calculated by:

    a = \frac12 \cdot c \cdot h ........ [1] or

    a = \frac12 \cdot a \cdot b ........ [2]

    and therefore: \frac12 \cdot c \cdot h = \frac12 \cdot a \cdot b........... Solve this equation for h and you'll get:

    \boxed{  h = \frac{a \cdot b}{c}}
    That is very smart
    It avoids similarity completely!!! I love such elegant ideas, thanks earboth
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