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prove that of all parallelogram of given perimeter , a rectangle has the greatest area 

Asked by sinha.nid25 24th February 2015, 6:16 PM
Answered by Expert
Answer:

 begin mathsize 14px style In space square to the power of straight m space ABCD space of space sides space straight a space and space straight b comma space let space straight h space be space the space height space corresponding space to space the space base space straight a. Now space for space increment DAE comma space straight b space is space the space hypotenuse. space We space know space that space straight a space hypotenuse space is space the space longest space side space of space straight a space right space angle space triangle. therefore straight h space less than space straight b space space space space space space space space space space space space space space space space space space space..... space left parenthesis 1 right parenthesis Now comma Area space of space straight a space parallelogram space equals space ah Next comma Area space of space rectangle space equals space ab Then space using space left parenthesis 1 right parenthesis space we space have comma straight h space less than space straight b therefore ah space less than space ab therefore space Area space of space parallelogram space less than space Area space of space Rectangle  Thus space straight a space rectangle space has space the space greatest space area space amongst space all space parallelograms. end style

Answered by Expert 24th February 2015, 6:34 PM
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