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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Application of Derivatives

NDA 2018 · Mathematics · Application of Derivatives

Paragraph: Three sides of a trapezium are each equal to 6 cm. Let α...

Paragraph: Three sides of a trapezium are each equal to 6 cm. Let \alpha \in \left(0, \frac{\pi}{2}\right)be the angle between a pair of adjacent sides.
Question: If the area of the trapezium is the maximum possible, then what is \alpha equal to?

Options

  1. Aπ6\frac{\pi}{6}
  2. Bπ4\frac{\pi}{4}
  3. Cπ3\frac{\pi}{3}
  4. D2π5\frac{2 \pi}{5}

Correct answer

C. π3\frac{\pi}{3}

Step-by-step solution

Length of three sides of trapezium =6cm=6 \mathrm{ cm} Let A Ebeheightoftrapezium.<br/>be height of trapezium. <br/> In △ADE,AD2=AE2+DE2⇒62=AE2+x2⇒AE2=36−x2⇒AE=36−x2\begin{aligned} & \text{ In } \triangle\mathrm{ADE}, \mathrm{AD}^2=\mathrm{AE}^2+\mathrm{DE}^2 \\ & \Rightarrow 6^2=\mathrm{AE}^2+\mathrm{x}^2 \\ & \Rightarrow \mathrm{AE}^2=36-\mathrm{x}^2 \\ & \Rightarrow \mathrm{AE}=\sqrt{36-\mathrm{x}^2}\end{aligned}AreaofArea oftrapezium =\frac{1}{2} h(a+b)=1236−x2(6+6+2x)A=(6+x)36−x2\begin{aligned} & =\frac{1}{2} \sqrt{36-x^2}(6+6+2 x) \\ & A=(6+x) \sqrt{36-x^2} \end{aligned}Given,Areaoftrapeziumismaximum.Given, Area of trapezium is maximum.dAdx=ddx[(6+x)⋅36−x2]=(6+x)⋅(−2x236−x2)+36−x2\begin{aligned} & \frac{d A}{d x}=\frac{d}{d x}\left[(6+x) \cdot \sqrt{36-x^2}\right] \\ & =(6+x) \cdot\left(\frac{-2 x}{2 \sqrt{36-x^2}}\right)+\sqrt{36-x^2} \end{aligned}$\begin{aligned} & =\sqrt{36-x^2}-\frac{x(6+x)}{\sqrt{36-x^2}} \\ & =\frac{36-\mathrm{x}^2-6 \mathrm{x}-\mathrm{x}^2}{\sqrt{36-\mathrm{x}^2}}=\frac{36-6 \mathrm{x}-2 \mathrm{x}^2}{\sqrt{36-2 \mathrm{x}^2}} \\ & \frac{\mathrm{dA}}{\mathrm{dx}}=0 \Rightarrow 36-6 \mathrm{x}-2 \mathrm{x}^2=0 \\ & \Rightarrow 2 \mathrm{x}^2+6 \mathrm{x}-36=0 \\ & \Rightarrow \mathrm{x}^2+3 \mathrm{x}-18=0 \\ & \Rightarrow \mathrm{x}^2+6 \mathrm{x}-3 \mathrm{x}-18=0 \\ & \Rightarrow \mathrm{x}(\mathrm{x}+6)-3(\mathrm{x}+6)=0 \\ & (x+6)(x-3)=0 \end{aligned}\Rightarrow \quad x=-6or3.Sinceor 3 . Sincexcannotbenegative.So,cannot be negative. So,x=3..\thereforeInIn\triangle \mathrm{ADE}, \cos \alpha=\frac{\mathrm{x}}{6}=\frac{3}{6}=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}$.
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