Question

please answer the question

Answer 1

triangle PQR is right angle triangle .
so , use Pythagoras theorem ,

PR^2 =PQ^2 +QR^2

area of triangle = 1/2 height x base
area of PQR = 1/2 PQ x QR

PQ.QR =2a ----------------(1)

but QR = b
and PQ ={ PR^2 - QR^2 }^1/2
={ PR^2 - b^2 }^1/2
also area of PQR = a

hence ,
a = 1/2 x { PR ^2 - b^2 }^1/2 x b

take square both side ,
a^2 = 1/4 (PR^2 -b^2 ) b^2
4a^2/b^2 =PR^2 - b^2
4a^2/b^2 + b^2 = PR^2
(4a^2 + b^4)/b^2 =PR^2
take square root
PR = (4a^2 + b^4 )^1/2/b ------------(2)

now ,
question ask ,
QN perpendicular PR
triangle PQN and triangle PQR
angle QPN = angle QPR
angle PNQ = angle PQR
so ,
triangle PNQ ~ triangle PQR
so,
PQ/PR = QN /QR =PN/PQ
so,
PQ /PR = QN/QR
QN =(PQ/PR) QR
={PQ. QR}/PR
now equation (1) and (2)

QN = 2ab/{4a^2 + b^4 }^1/2
hence proved .

Answer 2

ΔPQR is right angled at Q.
QR = b
a = Area(ΔPQR) = PQ * QR /2 =  PQ * b /2
        =>  PQ = 2a / b

Using Pythagoras theorem,  
       PR² = PQ² + QR² = 4a²/b²  + b² = (4a² + b⁴)/b²

QN ⊥ PR.  
     So  Area(ΔPQR) = 1/2 * PR * QN = QN * √(4a²+b⁴) / 2b

Given  Area(ΔPQR) = a = QN * √(4a²+b⁴) / 2b

=>  QN = 2ab / √(4a² + b⁴)