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Question
The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that:
AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2.4 m. The tunnel is 40 m long. Calculate:
(i) The cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per m2 (sq. meter).
(ii) The cost of paving the floor at the rate of Rs. 18 per m2.
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Solution
The cross-section of a tunnel is of the trapezium-shaped ABCD in which AB = 7 m, CD = 5 m and AM = BN. The height is 2.4 m and its length is 40 m.
(i) AM = BN =`( 7 - 5 )/( 2 )= ( 2 )/( 2 ) =1"m"`
∴ In ΔADM,
AD2 = AM2 + DM2 ...[ Using Pythagoras theorem ]
= 12 + (2 . 4)2
= 1 + 5.76
= `sqrt6.76`
= 2.6
AD = 2.6 m
Perimeter of the cross-section of the tunnel = ( 7 + 2.6 + 2.6 + 5 ) m = 17.2 m
Length = 40 m
∴ The internal surface area of the tunnel ( except the floor )
= ( 17.2 × 40 - 40 × 7) m2
= ( 688 - 280 ) m2
= 408 m2
Rate of painting = Rs. 5 per m2
Hence, total cost of painting = Rs. 5 × 408 = Rs. 2040
(ii) Area of floor of tunnel = l × b = 40 × 7 = 280 m2
Rate of cost of paving = Rs. 18 per m2
Total cost = 280 × 18 = Rs. 5040
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