AlexandervonHumboldt2
  • AlexandervonHumboldt2
Segments, joining the midpoints of opposite sides (A_1, B_1, C_1, D_1) of the quadriletiral ABCD are perpendicular and are equal to 2 and 7. Find the area of ABCD. So far i had done that A_1B_1C_1D_1 is a phombus.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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AlexandervonHumboldt2
  • AlexandervonHumboldt2
@Michele_Laino @hartnn @Hero @IrishBoy123 @amistre64
AlexandervonHumboldt2
  • AlexandervonHumboldt2
|dw:1447009845509:dw|
amistre64
  • amistre64
so a1 to b1 isnt a segment being defined? the segments join the opposite midpoints?

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AlexandervonHumboldt2
  • AlexandervonHumboldt2
woops
amistre64
  • amistre64
|dw:1447010080786:dw| what are 2 and 7 defining?
AlexandervonHumboldt2
  • AlexandervonHumboldt2
no sorry
AlexandervonHumboldt2
  • AlexandervonHumboldt2
A_1C_1=7 D_1B_1=2 sorry got confused in my own drawing
amistre64
  • amistre64
|dw:1447010258341:dw|
amistre64
  • amistre64
i got the 2 and 7 in the wrong spots tho
amistre64
  • amistre64
and the points ABCD are otherwise unknown correct?
AlexandervonHumboldt2
  • AlexandervonHumboldt2
yeah they are random
amistre64
  • amistre64
|dw:1447010697550:dw| hmm,im not privy to alot of geometry thrms in my head, and im not sure if determining the side lengths would be useful,other than converting it to a coordinant system. 13.25 = a^2 +b^2 -2ab cos(t1) b^2 +k^2 -2bk cos(t2) k^2 +g^2 -2kg cos(t3) g^2 +a^2 -2ga cos(t4)
amistre64
  • amistre64
not even sure if they can have specifics, but maybe defining a specific set would help determine a general approach
AlexandervonHumboldt2
  • AlexandervonHumboldt2
i'll think about it
amistre64
  • amistre64
|dw:1447011152704:dw| for a specific example,we could try working a trapezoid, that would give the same setup with something that is known
Michele_Laino
  • Michele_Laino
I tried to draw your shape, and I didn't get a rhombus @AlexandervonHumboldt2
AlexandervonHumboldt2
  • AlexandervonHumboldt2
hmmmm.
AlexandervonHumboldt2
  • AlexandervonHumboldt2
guys i found how to solve it! will write in a sec
Michele_Laino
  • Michele_Laino
The perpendicularity is part of the hypothesis
Michele_Laino
  • Michele_Laino
I think that the requested area is 14
AlexandervonHumboldt2
  • AlexandervonHumboldt2
Let K, L M and N be midpoints of AB, BC, CD, AD where LN=2 and KM=7. KL and MN are midlines of triangles ABC and ADC. Thus KL || AC, KL=1/2 AC MN=1/2 AC, MN || AC. => KLMN-parallelogram. As KL is the midline of ABC the area of KBL is 1/4 of ABC and area of MDN is 1/4 of ADC. Thus S_KBL+S_MDN=1/4 S_ABC+ /4 S_ADC = 1/4 S_ ABCD S_KAN+S_MCL=1/4 S_ABCD S_KLMN=S_ABCD-S_KBL - S_MDN- S_KAN- S_MCL=1/2 S_ABCD S_ABCD=2*KLMN=2*7=14
AlexandervonHumboldt2
  • AlexandervonHumboldt2
lol 14 is answer
AlexandervonHumboldt2
  • AlexandervonHumboldt2
you are right @Michele_Laino and thanks @amistre64
Michele_Laino
  • Michele_Laino
here is my reasoning: |dw:1447011964698:dw| we can write this equations: \(a^2+b^2=a^2+d^2\) so we have \(b=d\) furthermore, we have: \(a^2+d^2=d^2+c^2\), so we have \(c=a\)
AlexandervonHumboldt2
  • AlexandervonHumboldt2
nice reasoning :)
Michele_Laino
  • Michele_Laino
:)
Michele_Laino
  • Michele_Laino
I have based my reasoning on the observation that \(A_1,B_1,C_1D_1\) is a rhombus
amistre64
  • amistre64
ax = -bx (ay+by)/2 =1 cx = -dx (cy+dy)/2 =-1 (ax+dx)/2 = -3.5 ay= -dy (bx+cx)/2 = 3.5 by = -cy -------------------- L1 = (by+dy)/(2bx) (x-bx) + by L2 = (dy+by)/(2dx) (x-dx) + dy L3 = 2by/(bx+dx)(x-bx) + by L4 = 2dy/(bx+dx)(x-dx)+ dy -------------------- first thought was integration :) otherwise, we can just add up areas of the appropriate parts.
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