anonymous
  • anonymous
Geometry Help! Please!
Mathematics
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anonymous
  • anonymous
Geometry Help! Please!
Mathematics
jamiebookeater
  • jamiebookeater
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welshfella
  • welshfella
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
welshfella
  • welshfella
area= BE * DE - x y = 2* (3 - x) y = 6 - 2x
welshfella
  • welshfella
which is the same equation as we had before! how embarrassing!! I must be missing something here.....

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welshfella
  • welshfella
the area is a rhombus - so is there a formula for this in terms of the sides - or diagonals?
ganeshie8
  • ganeshie8
Yes, here is an outline of solution step1 : show that all the four triangles outside the shaded region are congruent |dw:1440259209437:dw|
ganeshie8
  • ganeshie8
step2 : By CPCTC, it follows that BG = BH = DH = DG, making BGDH a rhombus
amistre64
  • amistre64
|dw:1440259500328:dw|
amistre64
  • amistre64
solve for x x^2 + 4 = (3-x)^2
ganeshie8
  • ganeshie8
step3 : \(\triangle OBG\sim\triangle ADB \) by \(AA\) similarity, so \(\dfrac{OG}{OB}=\dfrac{2}{3}\) |dw:1440259468027:dw|
ganeshie8
  • ganeshie8
you can find OB trivially using the fact that OB = BD/2 (diagonals bisect each other in rhombus)
welshfella
  • welshfella
yes
welshfella
  • welshfella
and the area of the rhombus = pq/2 where p and q are the diagonals right?
ganeshie8
  • ganeshie8
yes, otherwise it is easy to see that all the four internal triangles are congruent : |dw:1440260040970:dw|
ganeshie8
  • ganeshie8
\(A_1 = A_2 = A_3 = A_4\)
Loser66
  • Loser66
Is it not that it is 6 -2x And xis calculated by pythagorean, x=5รท6
Loser66
  • Loser66
Or I missed something?
ganeshie8
  • ganeshie8
you might be correct, but i like similar triangles more :)
welshfella
  • welshfella
amistre's method is correct also simple really
Loser66
  • Loser66
Of course we need prove the 2 triangle are equal first
ganeshie8
  • ganeshie8
why do you want to solve a quadratic when you can setup a simple proportion ?
amistre64
  • amistre64
lol, its not a quadratic, its linear
anonymous
  • anonymous
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.\[A = bh = (3-\frac{5}{6})2 = \frac{13}{3}\]
welshfella
  • welshfella
once you know the area of the triangles thats it Yes the congruency proff is necessary
ganeshie8
  • ganeshie8
it looked like a quadratic to me haha
welshfella
  • welshfella
yea - but the x^2 cancel out
anonymous
  • anonymous
Oh i get it now!
ganeshie8
  • ganeshie8
i mean my brain felt it is a quadratic, so rejected initially ;p
anonymous
  • anonymous
you all are so helpful thanks so much!
welshfella
  • welshfella
yw
amistre64
  • amistre64
one thing i aint, is a geometrer ;)
ganeshie8
  • ganeshie8
looking at it again, i find it is simpler and kinda cute !
welshfella
  • welshfella
interesting question...

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