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Sam ji

· started a discussion

· 1 Months ago

Pls explain. (Question: 111047.0 )

Pls explain how the area of semi-circular top has been calculated in the given solution by u. Rather explain the solution elaborately.

Thanx.

Question:
The cross-section of a railway tunnel is a rectangle 6m broad and 8m high, surounded by a semi-circle as shown in the adjoining figure. The tunnel is 35 m long. The cost of plastering the internal surface of the tunnel excluding the floor, at the rate of \(\unicode{x20B9} \)3 per m2 is :


Options:
A) \(\unicode{x20B9} \)2380
B) \(\unicode{x20B9} \)1780
C) \(\unicode{x20B9} \)2170
D) \(\unicode{x20B9} \)2670
Solution:
Ans: (d) (Area of parallel walls of height 8 m) + (Area of semi cylinderical top)

\(\therefore\) Area of wall \(=2\times(8\times35)=560m^2\)

\(\therefore\) Area of semi-cylinderical top = \(\frac{2\pi rh}{2}\)

\(=\cfrac{22}{7}\times3\times35=22\times3\times5=330\ m^2\)

\(\therefore\) Total area to be plastering \(=560+330=890m^2\)

\(\therefore\) Cost of plastering \(=\unicode{x20B9}(890\times3)=\unicode{x20B9}2670 \)

Knowledge Expert

· commented

· 1 Months ago

Dear student
Given answer is correct
In solution
Curved surface area of semi cylinder = 2pi*r*h / 2 = pi*r*h
And follow solution.

Best wishes
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