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Diagonal of a parallelogram


Big Dog 04

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hi bigdog,

 

first consider the point where the rt. triangle is formed as E.

 

Since its a parallelogram BC=AD=12.

 

draw a straight line from point C to AD and name it as F.

 

so, DF=CE=2 and CF=ED=4

 

so now u have AF=10, CF=4 use pythogaraeus theorem on triangle AFC ..

 

u get AC as sqrt(116) = 2sqrt(29)

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Mansi please open the attachment I have uploaded. It'll clear all your doubts. (ale ka laxat ata tari ?) :grad:

 

parag,

I solved this problem by dropping an extra perpendicular to get the ans. as

2sqrt(29) exactly like the way you did. But i dont seem to get the same ans. applying AB^2 + BC^2 + CD^2 + DA^2=AC^2 + BD^2....the ans. is 2sqrt(41).

 

 

can you clarify the reason for the two different answers ,commentor?

 

 

p.s: parag,what do you mean by "ale ka laxat ata tari ?"

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parag,

I solved this problem by dropping an extra perpendicular to get the ans. as

2sqrt(29) exactly like the way you did. But i dont seem to get the same ans. applying AB^2 + BC^2 + CD^2 + DA^2=AC^2 + BD^2....the ans. is 2sqrt(41).

 

 

can you clarify the reason for the two different answers ,commentor?

 

 

p.s: parag,what do you mean by "ale ka laxat ata tari ?"

 

I request you to open the attachment I have uploaded in my solution. Please refer to it so no doubts could persist.

 

PS : "ale ka laxat" means, did you undertand now ???

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