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Suppose the radius of semicircle, length and breadth of rectangle be r, x and y respectively
AE = r
AB = x = 2r(semicircle is mounted over rectangle) …1
AD = y
Given as the perimeter of window = 10 m
x + 2y + πr = 10
2r + 2y + πr = 10
2y = 10 – (π + 2).r
y = (10 - (π + 2)r)/2 ...(2)
To admit the maximum amount of light, area of the window should be maximum
Assuming the area of window as A
![](https://www.sarthaks.com/?qa=blob&qa_blobid=18316215843076797062)
For the r = 10/(4 + π) A will be the maximum.
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