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QS = SR, QU = SU, PW = WS, ST || RV

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by sanju09 » Sat Jan 09, 2010 1:45 am
In the following figure QS = SR, QU = SU, PW = WS, ST || RV


The area of ∆PSX/area of ∆PQR is equal to
(A) 1/5
(B) 1/7
(C) 1/6
(D) 1/9
(E) None of these
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by rohan_vus » Sat Jan 09, 2010 7:09 am
Should be 1/5..

Area of triangle PUS/Area of triangle PQR = 1/4..eqn .(1).....As height remains same but the base of PUS ( SU = 1/4 of QR based on conditions given)

Al we want to find is Ratio of Area of triangle PSX to Area of triangle PQR..

Now we if get how area of PSX is related to area of PUS then thats it.

Condider line PU..its divided in such a way that .. PY = YX and XU = PY/2....This you get as parallel lines divide the adjacent lines in same ratio..
( TQ is divided in half if you draw a line from point U towards PQ and parallel to TS ..Since TQ = VT = VP ..so you know how XU is related to line PU ..Simple similar triangle property gives this)

So we get 5*Area of PSX = 4*Area of PUS..

Thus original ratio as asked now becomes = 1/5

PS : I did it in rough drawing but cant put my rough work as image to show this so trying best to put in words
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by anugaur12356 » Sun Jul 07, 2019 1:55 am
plz expalin clearly with figure
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