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.

Answer 1

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