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In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the non-parallel sides are all equal to $30$ In order to maximize the area of the trapezium, the smallest angle should be.
A.$\dfrac{\pi }{6}$
B.$\dfrac{\pi }{4}$
C.$\dfrac{\pi }{3}$
D.$\dfrac{\pi }{2}$

Last updated date: 15th Jul 2024
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Hint: since we are given length of parallel sides and hence we should use
the area formula of parallel side and use it.
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Let the angle made by hypotenuse of the right angle triangle with the base the trapezium be $\theta $
now the length of the base of the triangle is $a\cos \theta $
and the length of the perpendicular of the triangle or the distance between the
parallel sides of the trapezium is $a\sin \theta $
now the area of the trapezium = $\dfrac{1}{2}$(sum of parallel sides)$ \times $(distance between the parallel sides)
now for our given trapezium substituting the values we get
area = A=$\dfrac{1}{2} \times (a + 2a\;cos\theta + a) \times (a\;sin\theta )$
where $a = 30$
now on solving A =\[{a^2} \times (\;cos\theta + 1) \times (sin\theta )\]
on opening the brackets and using the identity $\sin 2\theta = 2\cos \theta \sin \theta $
A=\[{a^2}(\dfrac{1}{2} \times sin2\theta + \;sin\theta )\]
Now to maximize A we need to differentiate the equation for A with respect
to $\theta {\text{ and put }}\dfrac{{dA}}{{d\theta }} = 0$
now on differentiating the equation of A we get and equating it to $0$
\[(\cos 2\theta + \;\cos \theta ) = 0\]
Now using identity of $\cos 2\theta = 2{\cos ^2}\theta - 1$
And simplifying we get
  2\;co{s^2}\theta + \;cos\theta - 1 = 0 \\
   \Rightarrow \cos \theta = \dfrac{1}{2}, - 1 \\
   \Rightarrow \theta = \dfrac{\pi }{3},\pi \\
But since we want the smallest angle hence answer is $\dfrac{\pi }{3}$,option C
Note: while calculating questions of area of various polygons use various trigonometric formulae to simplify the calculation