Weekend Geometry Puzzle.

Candid-B

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  • Apr 25, 2019
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    මේ විදිහට සම්භන්ධතාවයක් ගොඩනගන්න පුළුවන්ද ? වෘත්ත කේන්ද්‍රයේ ඉඳන් B වල දුර ගන්න. මගේ මොළේ කුරොවල් වෙලා වගේ නිතර වැල බලලා.
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    ------ Post added on Jul 14, 2024 at 11:00 AM
     
    Last edited:

    imhotep

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  • Mar 29, 2017
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    Thanks @jamiezue & @dilann for the solutions. @jamiezue , you made an error in the area of the triangle BCD. But very close to the actual.

    The simplest approach (possibly)... We immediately know the sides of the triangle BCD.
    BC = 13, BD = SQRT (18^2 + 6^2) = 6√10, CD = SQRT(6^2+5^2) = √61

    Now we have the three sides of the triangle. Moreover, we know the area of the triangle BCD which is (1/2) x 13 (base) x 6 (height) = 39 sq
    (Note - Heron's formula would have given us the same result, but no need that to be used as seen above)

    Now use the property of the circumscribed circle. R=abc/4Δ which straight away gives R = √610 / 2 = 12.35 (y)
     

    jamiezue

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  • Jul 28, 2008
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    -~උලක් උඩ-~
    Thanks @jamiezue & @dilann for the solutions. @jamiezue , you made an error in the area of the triangle BCD. But very close to the actual.

    The simplest approach (possibly)... We immediately know the sides of the triangle BCD.
    BC = 13, BD = SQRT (18^2 + 6^2) = 6√10, CD = SQRT(6^2+5^2) = √61

    Now we have the three sides of the triangle. Moreover, we know the area of the triangle BCD which is (1/2) x 13 (base) x 6 (height) = 39 sq
    (Note - Heron's formula would have given us the same result, but no need that to be used as seen above)

    Now use the property of the circumscribed circle. R=abc/4Δ which straight away gives R = √610 / 2 = 12.35 (y)
    Thank you!