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Explain solution RD Sharma class 12 Chapter 17 Maxmima and Minima Exercise Case Study Questions question 4 subquestion (v)

Answers (1)

Answer:  \frac{50}{\pi+4}

Hint: use formula of area of semicircle.

Solution:

Area of maximum length throughout the window  =\frac{\text { radius of Semicircle }(\mathrm{x})}{2}+\text { breadth of rectangle }

\begin{aligned} &=\left(\frac{20}{\frac{\pi+4}{2}}\right)+\frac{10}{\pi+4} \cdot \\ & \end{aligned}

=\frac{40}{\pi+4}+\frac{10}{\pi+4}=\frac{50}{\pi + 4}

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