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anonymous

  • one year ago

Geometry Help! Please!

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  1. welshfella
    • one year ago
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    The 4 triangles right angled formed are congruent Let the smallest side of these = x and let the area of shaded part equal y area of each triangle = 2* x/2 = x consider one of the large rectangles y + 2x = 6 consider the shaded part which is a paralelogram the area = base * perpendicular height

  2. welshfella
    • one year ago
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    area= BE * DE - x y = 2* (3 - x) y = 6 - 2x

  3. welshfella
    • one year ago
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    which is the same equation as we had before! how embarrassing!! I must be missing something here.....

  4. welshfella
    • one year ago
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    the area is a rhombus - so is there a formula for this in terms of the sides - or diagonals?

  5. ganeshie8
    • one year ago
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    Yes, here is an outline of solution step1 : show that all the four triangles outside the shaded region are congruent |dw:1440259209437:dw|

  6. ganeshie8
    • one year ago
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    step2 : By CPCTC, it follows that BG = BH = DH = DG, making BGDH a rhombus

  7. amistre64
    • one year ago
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    |dw:1440259500328:dw|

  8. amistre64
    • one year ago
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    solve for x x^2 + 4 = (3-x)^2

  9. ganeshie8
    • one year ago
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    step3 : \(\triangle OBG\sim\triangle ADB \) by \(AA\) similarity, so \(\dfrac{OG}{OB}=\dfrac{2}{3}\) |dw:1440259468027:dw|

  10. ganeshie8
    • one year ago
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    you can find OB trivially using the fact that OB = BD/2 (diagonals bisect each other in rhombus)

  11. welshfella
    • one year ago
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    yes

  12. welshfella
    • one year ago
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    and the area of the rhombus = pq/2 where p and q are the diagonals right?

  13. ganeshie8
    • one year ago
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    yes, otherwise it is easy to see that all the four internal triangles are congruent : |dw:1440260040970:dw|

  14. ganeshie8
    • one year ago
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    \(A_1 = A_2 = A_3 = A_4\)

  15. Loser66
    • one year ago
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    Is it not that it is 6 -2x And xis calculated by pythagorean, x=5÷6

  16. Loser66
    • one year ago
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    Or I missed something?

  17. ganeshie8
    • one year ago
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    you might be correct, but i like similar triangles more :)

  18. welshfella
    • one year ago
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    amistre's method is correct also simple really

  19. Loser66
    • one year ago
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    Of course we need prove the 2 triangle are equal first

  20. ganeshie8
    • one year ago
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    why do you want to solve a quadratic when you can setup a simple proportion ?

  21. amistre64
    • one year ago
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    lol, its not a quadratic, its linear

  22. anonymous
    • one year ago
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    Picking up from @amistre64 \[2^2 + x^2 = (3-x)^2\]\[x^2 + 4 = x^2-6x +9\]\[6x = 9-4\]\[x = \frac{ 5 }{ 6 }\]The figure is also a parallelogram with height 2 and length 3-5/6 = 13/6.\[A = bh = (3-\frac{5}{6})2 = \frac{13}{3}\]

  23. welshfella
    • one year ago
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    once you know the area of the triangles thats it Yes the congruency proff is necessary

  24. ganeshie8
    • one year ago
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    it looked like a quadratic to me haha

  25. welshfella
    • one year ago
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    yea - but the x^2 cancel out

  26. anonymous
    • one year ago
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    Oh i get it now!

  27. ganeshie8
    • one year ago
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    i mean my brain felt it is a quadratic, so rejected initially ;p

  28. anonymous
    • one year ago
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    you all are so helpful thanks so much!

  29. welshfella
    • one year ago
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    yw

  30. amistre64
    • one year ago
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    one thing i aint, is a geometrer ;)

  31. ganeshie8
    • one year ago
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    looking at it again, i find it is simpler and kinda cute !

  32. welshfella
    • one year ago
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    interesting question...

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