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Fool's problem of the day, A not so easy geometry problem, A triangle is divided into four parts by straight lines from two of the corners. The area of three triangular parts are 8,5 and 10 sq units. What is the area of the remaining part? Good luck!

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The diagram is not to scale.
does any of the line equal?

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Other answers:

i'd take it as a no
answer is 22
1 sec doesn't have the solution right?
oo thanks.... i am seeing this now!! wat u trying to prove ??
I am interested to see your solution :)
i m drwing it
Take your time :)
here it is assumed the base to be BC and the upper vertex to be A p is d pt. dropped frm vertex b at AC q is d pt. dropped frm vertex c at AB nd O is d pt. whr BP and CQ intersect each othr join PQ nw in tri. bpc areaBOC is twice of areaCOP so BO must b twice of PO and area of POQ cmes out to be 4(areaBOQ is 8) in tri.AQC(PQ is still joind) areaAQP/areaCPQ=AO/CO nd in tri.ABC areaABP/areaBPC=AO/CO THEREFORE(let areaAQP is X) areaAQP/areaCPQ=areaABP/areaBPC on solving areaAQP=18 so area of the remaining part in abc= 18+4=22
That is a short version of the above solution: Join the third vertex to the intersection of the two lines. Let x and y be the areas of the two triangles formed, with y near 8 and x near 5 \[ x = \frac 1 2 (y+8)\\ y= \frac 8 {10} (x+5) \] Solve, you get x=10, y= 12 and x+y =22

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