Question

$4(P{M}^2+R{N}^2)=5Q{R}^2$
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)
$R{N}^2=Q{R}^2+(\frac{PQ}{2}{)}^2$ (N is mid point of PQ)
$4R{N}^2=4Q{R}^2+P{Q}^2$...(I)
Now, In $△PQM$,
$P{M}^2=P{Q}^2+Q{M}^2$ (By Pythagoras theorem)
$P{M}^2=P{Q}^2+(\frac{QR}{2}{)}^2$ (M is mid point of QR)
$4P{M}^2=4P{Q}^2+Q{R}^2$...(II)
Adding I and II,
$4(R{N}^2+P{M}^2)=4P{Q}^2+4Q{R}^2+Q{R}^2+P{Q}^2$
$4(R{N}^2+P{M}^2)=5(P{Q}^2+Q{R}^2)$
In $△PQR$,
$P{R}^2=P{Q}^2+Q{R}^2$
Hence, $4(R{N}^2+P{M}^2)=5P{R}^2$