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Callisto

  • 3 years ago

Geometry problem #1

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  1. Callisto
    • 3 years ago
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  2. shubhamsrg
    • 3 years ago
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    let each side of sq = a total area = 5a^2 thus area of triangle NDM = (5/2)a^2 let FA and CD meet at X ,,BA and ED at Y,,and MX =x,,let NY =y now area of triangle NDM = area of YAXD + area of NYA + area of NXM => (3)a^2 = ay + ax =>y+x = 3a =>y^2 + x^2 + 2yx = 9a^2 we also see that triangles MAX and NAY are similar =>x/a = a/y =>xy = a^2 --> x^2 + y^2 = 7a^2 now (y+a)^2 + (x+a)^2 = MN^2 =>(y^2+x^2) + 2a^2 + 2a(3a) = MN^2 =>15a^2 = MN^2 MN = a*sqrt(15)

  3. Callisto
    • 3 years ago
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    now area of triangle NDM = area of YAXD + area of NYA + area of NXM => (3)a^2 = ay + ax =>(2.5)a^2 = ay + ax ?

  4. shubhamsrg
    • 3 years ago
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    well it is 2.5a^2 = a^2 + 1/2 ay + 1/2 ax simplify a bit and you'll what i wrote..

  5. Callisto
    • 3 years ago
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    I'm sorry... still need more time to catch up with your solution :|

  6. Callisto
    • 3 years ago
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    Got it! Thanks !!!! :)

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