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Square

In a rectangl...

Question

In a rectangle $A B C D$ as shown in figure a point $P$ is taken on the side $C D$ such that $P C = 9, B P = 15$ and $A B = 14,$ then the correct relation between angles of $Δ A P B$ is:

α > β > γ

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α > γ > β

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β > γ > α

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γ > α > β

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Solution

The correct option is A $α > β > γ$
In $△ P B C$ , using Pythagoras theorem,
$P B^{2} = P C^{2} + B C^{2}$
$15^{2} = 9^{2} + B C^{2}$
$B C = 12$
Now, $A D = B C = 12$
and, $P D = C D - P C = A B - P C = 14 - = 5$
In $△ A P D$
$A P^{2} = A D^{2} + P D^{2}$
$A P^{2} = 5^{2} + 12^{2}$
$A P = 13$ cm
Now, $P B > A B > A P$
or $α > β > γ$ (Angles opposite longer sides are bigger)

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