In an isosceles trapezium,the length of one o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the parallel sides be $a = 30$ and $b = 30 + 2(30 \cos 	heta) = 30 + 60 \cos 	heta$. The height of the trapezium is $h = 30 \sin 	heta$.The

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) Let the parallel sides be $a = 30$ and $b = 30 + 2(30 \cos 	heta) = 30 + 60 \cos 	heta$. The height of the trapezium is $h = 30 \sin 	heta$.The area $A$ of the trapezium is given by:$A = \frac{1}{2}(a + b)h = \frac{1}{2}(30 + 30 + 60 \cos 	heta)(30 \sin 	heta)$$A = \frac{1}{2}(60 + 60 \cos 	heta)(30 \sin 	heta) = 900(1 + \cos 	heta) \sin 	heta = 900(\sin 	heta + \sin 	heta \cos 	heta) = 900(\sin 	heta + \frac{1}{2} \sin 2	heta)$.To maximize the area,differentiate with respect to $	heta$ and set to $0$:$\frac{dA}{d	heta} = 900(\cos 	heta + \cos 2	heta) = 0$.Since $\cos 2	heta = 2 \cos^2 	heta - 1$,we have:$2 \cos^2 	heta + \cos 	heta - 1 = 0$.$(2 \cos 	heta - 1)(\cos 	heta + 1) = 0$.This gives $\cos 	heta = \frac{1}{2}$ or $\cos 	heta = -1$.Since $	heta$ is an angle in a triangle,$	heta = \frac{\pi}{3}$ (as $	heta = \pi$ is not possible).Thus,the smallest angle is $\frac{\pi}{3}$.

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:

Similar Questions

The curve $y(x) = ax^{3} + bx^{2} + cx + 5$ touches the $x$-axis at the point $P(-2, 0)$ and cuts the $y$-axis at the point $Q$,where the derivative $y'(0) = 3$. Find the local maximum value of $y(x)$.

Statement-$I$: Let the function $f(x) = \begin{cases} -\frac{x}{2} & x < 0 \\ 7x + 8 & x \geq 0 \end{cases}$. Then $f(x)$ has a local minimum at $x = 0$.
Statement-$II$: If $f(a) < f(a - h)$ and $f(a) < f(a + h)$ for a sufficiently small $h > 0$,then $f(x)$ has a local minimum at $x = a$.

The maximum value of $(1/x)^x$ is -

If ${a^2}{x^4} + {b^2}{y^4} = {c^6}$,then the maximum value of $xy$ is

$A$ square piece of tin of side $18 \, cm$ is to be made into a box without a top by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible (in $, cm$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module