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 ₹ 5 per m² (sq. metre).

(ii) the cost of paving the floor at the rate of ₹ 18 per m².

(i) Length of the tunnel = 40 m

Height of the tunnel = 2.4 m

Cross-section dimensions: AB = 7 m, CD = 5 m

AM = BN = $\dfrac{AB - CD}{2}$ = $\dfrac{7 - 5}{2}$ = $\dfrac{2}{2}$ = 1 m

In Δ ADM,

Using the pythagorean theorem in triangle ADM,

Base² + Height² = Hypotenuse²

⇒ AM² + DM² = AD²

⇒ 1² + (2.4)² = AD²

⇒ 1 + 5.76 = AD²

⇒ 6.76 = AD²

⇒ AD = $\sqrt{6.76}$

⇒ AD = 2.6 m

Area of the internal surface of the tunnel (excluding the floor) = Area of vertical walls - Area of floor

= 2.6 x 40 + 5 x 40 + 2.6 x 40 m²

= 104 + 200 + 104 m²

= 408 m²

Total cost of painting = Area of vertical walls x Cost of painting

= ₹ 408 x 5

= ₹ 2,040

Hence, total cost of painting the internal surface = ₹ 2,040.

(ii) Area of floor = 7 x 40 m²

= 280 m²

Total cost of paving = Area of floor x rate of paving

= ₹ (280 x 18)

= ₹ 5,040

Hence, total cost of paving = ₹ 5,040.