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Mathematics 12 Online
OpenStudy (shubhamsrg):

geometry + similarity question i'd guess.. PQRS is a //gm PS//ZX PZ / QZ = 2/3 find XY/SQ =?

OpenStudy (anonymous):

Is there a second parallelogram adjacent to the fist? I'm trying to imagine the scenario.

OpenStudy (shubhamsrg):

|dw:1350050559015:dw|

OpenStudy (anonymous):

Ah.

OpenStudy (anonymous):

I can start by stating the obvious: SX/QX = 2/3.

OpenStudy (shubhamsrg):

hmm,,yep.. here's what i tried to do.. |dw:1350051162014:dw|

OpenStudy (shubhamsrg):

will this be helpful ?

OpenStudy (anonymous):

YM // to XZ? Hmm, yes, possibly..

OpenStudy (shubhamsrg):

yep.. //

OpenStudy (shubhamsrg):

further,, we may also see.. |dw:1350051445715:dw| let PZ = 2l thenZQ = 3l let ZM = j then MQ = 3l -j

OpenStudy (shubhamsrg):

|dw:1350051539893:dw| we can easily see a/(a+b+c) = 2/5 b/(a+b+c)= j/5l c/(a+b+c) =(3l -j)/5l we need to find numerical value of b/(a+b+c) where, we know that a/(b+c) = 2/3 any help ?

OpenStudy (anonymous):

Is it 1/10

OpenStudy (shubhamsrg):

nopes..

OpenStudy (shubhamsrg):

nopes according to my book i mean,,i dont know! :P what was your solution ?

ganeshie8 (ganeshie8):

is it 1/3 ?

OpenStudy (anonymous):

It's a/5-a whatever that is...:-)

OpenStudy (shubhamsrg):

nopes..not that either..

OpenStudy (anonymous):

Here's what I got so far. I let PZ=2 and ZQ=3 since that satisfies the ratio. XQ/ZQ = SX/PZ -> XQ/3 = SX/2 XQ/ZQ = 2/3 -> XQ = 2 SX/PZ = XQ/ZQ -> SX/2 = 2/3 SX = 4/3 SX+XQ = 10/3

OpenStudy (shubhamsrg):

XQ/ZQ = SX/PZ ?? how sir ?

OpenStudy (anonymous):

Similar triangles.

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