Question

In ∆ PQR, ∠ PQR = 90°, seg QN is the median of seg PR. If QN = 10, Find PR.

Answer 1

Answer:

ΔPQR , Right angled at Q and QR = b and Area of ΔPQR = a

QN is perpendicular to PR.

To Prove :

length of QN = 2ab/√b^{4} + 4a^{2}b

4

+4a

2

Solution:

Consider ΔPQR

Area = \frac{1}{2}

2

1

PQ.QR = a

2a/QR = PQ

Therefore

PQ = 2a/b

Consider ∠R

tan ∠R = PQ/b = 2a/b²

Now consider ΔQNR

sin ∠R = QN/QR

QN = sin ∠R x QR = b x sin ∠R

sin∠R = opposite side / hypotenuse

sin ∠R = PQ/PR = \frac{\frac{2a}{b} }{\sqrt{b^{2} + (\frac{2a}{b} )^{2} } }

b

2

+(

b

2a

)

2

b

2a

= \frac{2a}{\sqrt{b^{4} + 4a^{2} } }

b

4

+4a

2

2a

Therefore

QN = b x sin

∠R = \frac{2ab}{\sqrt{b^{4} + 4a^{2} } }

b

4

+4a

2

2ab