shubhamsrg
what is the ratio of area of a traingle ABC to the area of the triangle whose sides are equal to the medians of ABC ?
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shubhamsrg
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or,,prove that this ratio always equals 4/3
mukushla
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im watching :)
estudier
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|dw:1350313254678:dw|
estudier
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x^2 + h^2 = c^2 and y^2 +h^2 = b^2, x+y = a
x^2-y^2 = c^2-b^2 ......hmm might be better off with Cosine Rule:-(
estudier
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Yukky algebra....think, think...
estudier
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|dw:1350318976709:dw|
estudier
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|dw:1350319309051:dw|
estudier
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|dw:1350319508249:dw|
estudier
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152 is a quarter 123
-> 123/235 is 4/3
-> 143/362 is 4/3 as well
estudier
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Think that's right...
shubhamsrg
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why is 352 half of 362 ?
estudier
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Median divides triangle area in half?
shubhamsrg
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how is 35 a median ?
ganeshie8
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|dw:1350323581495:dw|
ganeshie8
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72 || 34 coz 72 is the midsegment
ganeshie8
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|dw:1350323706822:dw|
ganeshie8
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7824 is a parallelog gram,
now imagine we translate it down
ganeshie8
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|dw:1350323784808:dw|
estudier
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Yes, sorry, I didn't make the construction very clear.......
ganeshie8
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it translates to 1672
shubhamsrg
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why will 87 meet 6 ?
ganeshie8
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diagonals bisect each other in ||gram
ganeshie8
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so 65=52
ganeshie8
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87 meets at 6, thats right, but before that we should prove this :
326 is the triangle made by medians
ganeshie8
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87 meets at 6 becoz,
47 is congruent to 26,
42 is congruent to 21
ganeshie8
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the only tricky part in this proof i see is proving this :
326 is the triangle made by medians
shubhamsrg
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i still dont get 876 thing..
ganeshie8
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ok did u get estudier diagram
shubhamsrg
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yep..i followed every bit till now..
ganeshie8
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he constructed,
26 \(\cong\) 47
shubhamsrg
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yes..
ganeshie8
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26 is just translation of 47, so they are ||
one more construction he did is, he translated other median also, the translated medians intersect at 6. (we can prove this, later)
shubhamsrg
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ahh alright,,so 36 isnt a median yet..okay,,
ganeshie8
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assume it is a median so that we prove the thing at hand, :)
shubhamsrg
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alright,,
ganeshie8
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72 || 84, agree ?
shubhamsrg
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yes..
ganeshie8
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becoz 72 is the midsegment
shubhamsrg
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yes,,i could figure out..
ganeshie8
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7284 is a ||gram i have constructed
shubhamsrg
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7248*
shubhamsrg
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yes,,clear till now..
ganeshie8
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also, 74 congruent to 62, by construction
shubhamsrg
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okay,,7426 is a //gm..
ganeshie8
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now imagine we slide the ||gram 7248 along the side 41
shubhamsrg
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so you're saying those 2 //gms are congruent..
ganeshie8
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the diagonal in the ||gram 7248 translated to where ?
shubhamsrg
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coincides with 62
ganeshie8
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|dw:1350324847401:dw|
ganeshie8
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the diagonal 74 coincides wid 62
the side 42 coincides wid 21
shubhamsrg
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yes,,the 2 //gms are congruent..
ganeshie8
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|dw:1350324939445:dw|
ganeshie8
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7248 congruent 6127
ganeshie8
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we just use this : 6127 is a ||gram
that gives 5 is midpoint of diagonal 62
shubhamsrg
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that one we already knew right ?
ganeshie8
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which one ? i thought we're trying to prove 5 is midpoint of 62 lol
ganeshie8
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whats ur question
shubhamsrg
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we knew 5 was the mid point quite earlier..after that assumption, everything fitted..those 2 became congruent //gms and everything else..
ganeshie8
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no
shubhamsrg
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??
ganeshie8
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those 2 became congruent || grams by construction, 5 got nothing to do wid it
ganeshie8
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let me put everything in one reply
shubhamsrg
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we assumed 36 to be the median..that gave 16 // 27 ..by which it was a //gm,, and thus 5 was the midpoint ( 2 diagnols intersected)
ganeshie8
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ok if that make sense good :)
ganeshie8
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so only assumption we need to prove is 36 is a median, right ?
shubhamsrg
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yep..
sauravshakya
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shubhamsrg
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excellency yet again..!! thank you @sauravshakya
shubhamsrg
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and big thank you to @ganeshie8 ..your idea was used here after all..
shubhamsrg
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+ @estudier you sparked the figure, ofcourse ! :P
estudier
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I managed to get an outline of this just with algebra --- what a mess!!