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a , b , c , d are natural number such that a = bc , b = cd , c = da , d = ab. Then (a+b) (b+c) (c+d) (d+a) is equal to : 1. (a+b+c+d)^2 2.(a+d)^2 + (c+d)^2 3.(a+d)^2 +(b+c)^2 4.(a+c)^2 +(b+d)^2

a , b , c , d are natural number such that a = bc , b = cd , c = da , d = ab. Then (a+b) (b+c) (c+d) (d+a) is equal to : 
1. (a+b+c+d)^2 
2.(a+d)^2 + (c+d)^2 
3.(a+d)^2 +(b+c)^2 
4.(a+c)^2 +(b+d)^2

Grade:9

1 Answers

dip
24 Points
6 years ago
a=bc
b=cd
c=da
d=ab
therefore multiplying lhs & rhs separately abcd=(abcd)^2
or abcd=1 as abcd not equal to 0
since a,b,c,d are natural nos and abcd=1
therefore a=1,b=1,c=1,d=1
therefore dy putting valus of a,b,c,d in expression and options we see that
option 1 is correct

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