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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!
 one year ago
 one year ago
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!
 one year ago
 one year ago

This Question is Closed

FoolForMathBest ResponseYou've already chosen the best response.0
dw:1338105988238:dw
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
The diagram is not to scale.
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
does any of the line equal?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
http://www.pagalguy.com/forum/quantitativequestionsandanswers/71240officialquantthreadcat2011a796.html doesn't have the solution right?
 one year ago

shivamsinhaBest ResponseYou've already chosen the best response.1
oo thanks.... i am seeing this now!! wat u trying to prove ??
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
I am interested to see your solution :)
 one year ago

shivamsinhaBest ResponseYou've already chosen the best response.1
here it is i.ve 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
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
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
 one year ago
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