The cross-section of a tunnel, perpendicular to its length is a trapezium ABCD in which AB = 8 m, DC = 6 m and AL = BM. The height of the tunnel is 2.4 m and its length is 40 m. Find :

(i) the cost of paving the floor of the tunnel at ₹ 16 per m².

(ii) the cost of painting the internal surface of the tunnel, excluding the floor at the rate of ₹ 5 per m².

Given,

AB = 8 m

DC = 6 m

Height = 2.4 m

Length of the tunnel = 40 m

AL = BM

From figure, LM = DC = 6 m (since perpendiculars drop from D and C).

From figure,

⇒ AB = AL + LM + MB

⇒ 8 = AL + 6 + BM

Since, AL = BM

∴ 8 = 2AL + 6

⇒ 2AL = 8 - 6

⇒ 2AL = 2

⇒ AL = $\dfrac{2}{2}$ = 1 m

∴ AL = BM = 1 m.

Calculating the lengths of the sloping sides AD and BC,

In right triangle ALD,

AL = 1 m

DL = 2.4 m

Using pythagoras theorem for the triangle ALD,

⇒ Hypotenuse² = Base² + Height²

⇒ AD² = AL² + DL²

⇒ AD² = 1² + (2.4)²

⇒ AD² = 1 + 5.76

⇒ AD² = 6.76

⇒ AD = $\sqrt{6.76}$

⇒ AD = 2.6 m.

∴ AD = BC = 2.6 m.

(i) Cost of paving the floor:

Area of tunnel = AB × length

= 8 × 40

= 320 m².

Total cost = Area × cost per m²

= 320 × 16

= ₹ 5,120.

Hence, cost of paving the floor = ₹ 5,120.

(ii) Cost of painting internal surface (excluding floor)

Area to be painted are :

Roof DC and Two sloping sides AD and BC.

Calculating the area of DC,

Area = length × DC

= 40 × 6 = 240 m².

Calculating the area of slope AD,

Area = length × AD

= 40 × 2.6 = 104 m².

Calculating the area of slope BC,

Area = Length × BC

= 40 × 2.6 = 104 m².

Total area = 240 + 104 + 104

= 448 m².

Cost rate = ₹ 5 per m²

Calculating the cost of painting the internal surface of the tunnel,

Total cost = Total area × Cost

= 448 × 5

= ₹ 2,240.

Hence, cost = ₹ 2,240.