hey this is the question no.9
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Hi ,
In rect PQRS,
PQ+QR = Length + Breadth = x+y = 7
PR=QS=diagonal = sqrt(x^2 + y^2 ) ...using pythagoras theorem
so
2(sqrt(x^2+y^2)) = 10
x^2 + y^2 = 25
now
(x+y)^2 = x^2 + y^2 + 2xy ...using this identity, expand it and substitute their respective values
(x+y)^2 = x^2 + y^2 + 2xy
49 = 25 + 2xy
xy = 12
area = 12
Hope it helped.
In rect PQRS,
PQ+QR = Length + Breadth = x+y = 7
PR=QS=diagonal = sqrt(x^2 + y^2 ) ...using pythagoras theorem
so
2(sqrt(x^2+y^2)) = 10
x^2 + y^2 = 25
now
(x+y)^2 = x^2 + y^2 + 2xy ...using this identity, expand it and substitute their respective values
(x+y)^2 = x^2 + y^2 + 2xy
49 = 25 + 2xy
xy = 12
area = 12
Hope it helped.
hetvi51:
hmm :)
Let me know if you got other doubts
I'll be happy to help you L))
:))
ya I have but first I will try by my own if can't able to solve then I will surely tell to u kk
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wow... thats a good habit haha
thx hahahhaah
welcome again :))
:) :) :) :)
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