Question

Question no. 28plzzzzz....

Answer 1

Answer:

Step-by-step explanation:

Let∆PQR BE A RIGHT ANGLE AT 'Q' SUCH THAT QR=x andA area of∆ PQR

DRAW QN PERPENDICULAR to PR

NOW,AREA OF ∆PQR

A=I/2*QR*PQ

A=1/2*x*PQ

PQ=2A/x......(1)

Now in ∆ pqr and pnq

Angle pnq=angle pqr

And,angle p is common

So, by AAcriteria ∆pqr~∆pnq

PQ/PR=NQ/QR.....(2)

APPLYING PYTHAGORAS THEROM IN ∆PQR

PQ^2+QR^2=PR^2

4A^2/x^2+x^2=PR^2

PR=√4A^2+x^4/√b^2

PR=√4A^2+x^4/x

From eq1 n 2 we get,

2A/x*PR=NQ/x

NQ=2A/PR

PUTTING VALUE OF PR

WE GET NQ=2Ab/√4A^2+b^4

Hence proved

HOPE U WILL GET