If A is the area of a right angled triangle where - Q -90- and b is the base of the - PQR- then- find the length of altitude QN on the hypotenuse of theA- 2 Ab - b 4-4 A 2B- None of theseC- 2 Ab - A 4-4 b 2D- A b-b4-4 A2

The correct option is A 2Ab/√(b^4+4A^2) QR = b A = Ar(Δ PQR) A=1/2× b× PQ PQ=2A/b.....(1) Δ PNQ ∼Δ PQR(AA) ⇒PQ/PR=NQ/QR.....(2) From Δ PQR PQ^2+QR^2=PR^2 4 ...

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Byju's Answer

Standard X

Mathematics

Distance between Two Points on the Same Coordinate Axes

If A is the a...

Question

If A is the area of a right angled triangle where $∠ Q = 90^{\circ}$ and b is the base of the $Δ$ PQR, then, find the length of altitude QN on the hypotenuse of the triangle.

$\frac{2 A b}{\sqrt{b^{4} + 4 A^{2}}}$

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$\frac{2 A b}{\sqrt{A^{4} + 4 b^{2}}}$

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$\frac{A b}{\sqrt{b^{4} + 4 A^{2}}}$

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$N o n e o f t h e s e$

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Solution

The correct option is A
$\frac{2 A b}{\sqrt{b^{4} + 4 A^{2}}}$

QR = b
A = Ar( $Δ$ PQR)
$A = \frac{1}{2} \times b \times P Q P Q = \frac{2 A}{b} . . . . . (1) Δ P N Q \sim Δ P Q R (A A) \Rightarrow \frac{P Q}{P R} = \frac{N Q}{Q R} . . . . . (2) F r o m Δ P Q R P Q^{2} + Q R^{2} = P R^{2} \frac{4 A^{2}}{b^{2}} + b^{2} = P R^{2} P R = \sqrt{\frac{4 A^{2} + b^{2}}{b^{2}}} = \sqrt{\frac{4 A^{2} + b^{2}}{b}}$
Equation (2) becomes
$\frac{2 A}{b \times P R} = \frac{N Q}{b} N Q = \frac{2 A}{P R} N Q = \frac{2 A b}{\sqrt{4 A^{2} + b^{2}}}$

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If A is the area of a right angled triangle where ∠Q=90∘ and b is the base of the ΔPQR, then, find the length of altitude QN on the hypotenuse of the triangle.

The correct option is A 2Ab√b4+4A2 QR = b A = Ar(Δ PQR) A=12×b×PQPQ=2Ab.....(1)ΔPNQ∼ΔPQR(AA)⇒PQPR=NQQR.....(2)From ΔPQRPQ2+QR2=PR24A2b2+b2=PR2PR=√4A2+b2b2=√4A2+b2b Equation (2) becomes 2Ab×PR=NQbNQ=2APRNQ=2Ab√4A2+b2

If A is the area of a right angled triangle where $∠ Q = 90^{\circ}$ and b is the base of the $Δ$ PQR, then, find the length of altitude QN on the hypotenuse of the triangle.