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FoolForMath
Group Title
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!
 2 years ago
 2 years ago
FoolForMath Group Title
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!
 2 years ago
 2 years ago

This Question is Closed

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
dw:1338105988238:dw
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
The diagram is not to scale.
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
does any of the line equal?
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
i'd take it as a no
 2 years ago

shivamsinha Group TitleBest ResponseYou've already chosen the best response.1
answer is 22
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Explain~
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
http://www.pagalguy.com/forum/quantitativequestionsandanswers/71240officialquantthreadcat2011a796.html doesn't have the solution right?
 2 years ago

shivamsinha Group TitleBest ResponseYou've already chosen the best response.1
oo thanks.... i am seeing this now!! wat u trying to prove ??
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
I am interested to see your solution :)
 2 years ago

shivamsinha Group TitleBest ResponseYou've already chosen the best response.1
i m drwing it
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Take your time :)
 2 years ago

shivamsinha Group TitleBest 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
 2 years ago

eliassaab Group TitleBest 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
 2 years ago
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