are you sure you typed it correctly. as $a^2+b^2+c^2+d^2=ab+bc+cd+da \Rightarrow (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=0 $ or $a=b=c=d$. edit = hence quad. must be rhombus now $area = d1.d2/2 $ which gives $d2=4$.

aditya21 wrote: are you sure you typed it correctly. as $a^2+b^2+c^2+d^2=ab+bc+cd+da \Rightarrow (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=0 $ or $a=b=c=d$. thus $a^2=60$ but than we doesnt get length of diagonal as $30$. Try rechecking your logic.

aditya21 wrote: are you sure you typed it correctly. as $a^2+b^2+c^2+d^2=ab+bc+cd+da \Rightarrow (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=0 $ or $a=b=c=d$. thus $a^2=60$ but than we doesnt get length of diagonal as $30$. I think this means that it's a rhombus; by the diagonal formula for area, the other diagonal should be $\boxed{4}$...

yeah i was stupid to think it as square.

And algebra problem disguised as Geometry $a^2+b^2+c^2+d^2=ab+bc+cd+da\Rightarrow a=b=c=d$ by AM-GM inequality. Thus, it is a rhombus. Area of rhombus is $\frac{d_1d_2}{2}$ So, the other diagonal has length 4

YESMAths wrote: Let $ABCD$ be a convex quadrilateral with $AB=a$, $BC=b$, $CD=c$ and $DA=d$. Suppose \[a^2+b^2+c^2+d^2=ab+bc+cd+da,\]and the area of $ABCD$ is $60$ sq. units. If the length of one of the diagonals is $30$ units, determine the length of the other diagonal. Simple.. we may also use rearrangement inequality for this one.. a one-liner !