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Standard XII

Physics

Vector Addition

The line x+...

Question

The line $x + y = p$ meets the x-and y-axes at $A$ and $B$ , respectively. A triangle $△ A P Q$ is inscribed in the $△ O A B$ , $O$ being the origin, with right angle at $Q$ . $P$ and $Q$ lie, respectively, on $O B$ and $A B$ . If the area of the $△ A P Q$ is ${\frac{3}{8}}^{t h}$ of the area of the $△ O A B$ , then $\frac{A Q}{B Q}$ is equal to

2

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\frac{2}{3}

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\frac{1}{3}

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3

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Solution

The correct option is D $3$
$\frac{a r A Q P}{a r A O B} = \frac{3}{8}$ ...Given
Let $\frac{A Q}{B Q} = k$
Therefore, $\frac{\frac{p^{2} k}{(k + 1)^{2}}}{\frac{p^{2}}{2}} = \frac{3}{8}$
$\frac{2 k}{(k + 1)^{2}} = \frac{3}{8}$
$16 k = 3 k^{2} + 6 k + 3$
$3 k^{2} - 10 k + 3 = 0$
$k = \frac{1}{3}$ and $3$
But 3 is taken since $\frac{1}{3}$ gives a negative value.

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