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Standard VI

Mathematics

Area of Rectangle

In a rectangl...

Question

In a rectangle ABCD, M and N are the midpoints of the sides BC and CD respectively. If $ˆ A N M = 90^{0}$ then AB:BC is

2:1

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\sqrt{2} : 1

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\sqrt{3} : 1

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1:1

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Solution

The correct option is C $\sqrt{2} : 1$
In $Δ M C N, t a n θ = \frac{y}{x}$ and In $Δ A D N, t a n (90 - 0) = \frac{2 y}{x}$
$c o t θ = \frac{2 y}{x}$
$t a n θ = \frac{x}{2 y}$
$2 y^{2} = x^{2}$
$\frac{x}{y} = \frac{\sqrt{2}}{1}$

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The correct option is C $\sqrt{2} : 1$
In $Δ M C N, t a n θ = \frac{y}{x}$ and In $Δ A D N, t a n (90 - 0) = \frac{2 y}{x}$
$c o t θ = \frac{2 y}{x}$
$t a n θ = \frac{x}{2 y}$
$2 y^{2} = x^{2}$
$\frac{x}{y} = \frac{\sqrt{2}}{1}$