Question

triangle PQR is a right angled triangle .Right angle at Q such that QR=b and ,
a=A(triangle PQR).
If QN is perpendicular to PR
then show that QN=\frac{2ab}{ \sqrt{ {b}^{4} } + 4 {a}^{2} }

​

Answer 1

Given:

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

QN is perpendicular to PR.

To Prove :

length of QN = 2ab/√$b^{4} + 4a^{2}$

Solution:

Consider ΔPQR

• Area = $\frac{1}{2}$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 = $\frac{\frac{2a}{b} }{\sqrt{b^{2} + (\frac{2a}{b} )^{2} } }$ =  $\frac{2a}{\sqrt{b^{4} + 4a^{2} } }$

Therefore

• QN = b x sin∠R = $\frac{2ab}{\sqrt{b^{4} + 4a^{2} } }$

Answer 2

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