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Questions  >>  Olympiad-Math  >>  Class 10
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Q)

In the given figure, AP is equal to

$\begin{array}{1 1} 7\;cm \\ 6\;cm \\ 5.5 \;cm \\ 5\;cm \end{array} $

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A)
Solution :
In triangles APB and CPD,
$\angle APB = \angle CPD$ [Vertically opposite angles]
$\angle BAP = \angle ACD $ [Alternate angles as $AB \parallel CD$
$\Delta APB \approx \Delta CPD$ [AA similarity]
$\large\frac{AB}{CD} =\frac{CP}{AP}$
$\large\frac{4}{6} =\frac{AP}{7.5}$
$AP= \large\frac{7.5 \times 4 }{6}$$ = 5 \;cm$
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