If a,b,c,d are positive numbers, then ab+cd<2√abcd
True
False
As a,b,c,d are positive numbers, So ab,cd are also positive
Now as AM > GM
ab+cd2>2√(ab)(cd)
⇒ab+cd>2 √abcd
Hence given statement is False.
In a trapezium ABCD, if AB || CD then (AC2+BD2) = ?
(a) BC2+AD2+2BC.AD
(b) AB2+CD2+2AB.CD
(c) AB2+CD2+2AD.BC
(d) BC2+AD2+2AB.CD
If AB and CD are a pair of opposite sides of parallelogram ABCD, then