Question

$P{M}^2+R{N}^2=5M{N}^2$
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given : PQR is right triangle at Q. M and N are mid points of QR and PQ respectively.
Now, In $△NQR$,
$R{N}^2=Q{R}^2+N{Q}^2$ (By Pythagoras theorem)
$R{N}^2=(2QM{)}^2+N{Q}^2$ (M is mid point of QR)
$R{N}^2=4Q{M}^2+N{Q}^2$...(I)
Now, In $△PQM$,
$P{M}^2=P{Q}^2+Q{M}^2$ (By Pythagoras theorem)
$P{M}^2=(2QN{)}^2+Q{M}^2$ (N is mid point of PQ)
$P{M}^2=4Q{N}^2+Q{M}^2$...(II)
Adding I and II,
$R{N}^2+P{M}^2=4Q{M}^2+Q{N}^2+4Q{N}^2+Q{M}^2$
$R{N}^2+P{M}^2=5(Q{M}^2+Q{N}^2)$
In $△MQN$,
$M{N}^2=N{Q}^2+Q{M}^2$ (Pythagoras Theorem)
Hence, $R{N}^2+P{M}^2=5M{N}^2$