Mathematics - Straight Lines Question with Solution | TestHub

Let BD be integer and $l\ge m.2(l+m)=40\Rightarrow l+m=20$. Possible values of $m=1,2,3,\dots ,10$. Note: in any triplet of $l,m,BD$ if at least one is different, the parallelogram will be noncongruent. Now $l-m<BD<l+m$ (triangle inequality) $\Rightarrow 20-2m<BD<20\Rightarrow$ No. of possible values of BD for a given $m$ is $20-(20-2m)-1=2m-1\Rightarrow$ Total no. of noncongruent parallelograms $={\sum }_{m=1}^{10}(2m-1)=10^2=100$.