The cross-section of a tunnel is a square of side 7 m surmounted by a semicircle as shown in the adjoining figure. The tunnel is 80 m long. Calculate:

(i) its volume,

(ii) the surface area of the tunnel (excluding the floor) and

(iii) its floor area.

Given,

Side of square (a) = 7 m

Let radius of semi-circle be r metres.

⇒ 2r = 7

⇒ r = $\dfrac{7}{2}$ m.

Length of the tunnel (h) = 80 m

Area of cross section of the front part = Area of square + Area of semi-circle

= a² + $\dfrac{1}{2}πr^2$

= 7 x 7 + $\dfrac{1}{2} \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2}$

= 49 + $\dfrac{77}{4}$

= $\dfrac{196 + 77}{4} = \dfrac{273}{4}$ m².

(i) By formula,

Volume of the tunnel = Area of cross section x length of the tunnel

= $\dfrac{273}{4}$ x 80

= 5460 m³.

Hence, volume of tunnel = 5460 m³.

(ii) Surface area of the tunnel (excluding the floor) =

= Surface area of upper semi-circle portion + Surface area of square portion

= πrh + ah + ah

= $\dfrac{22}{7} \times \dfrac{7}{2} \times 80 + 7 \times 80 + 7 \times 80$

= 880 + 560 + 560

= 2000 m².

Hence, the surface area of the tunnel = 2000 m².

(iii) Area of floor = Breadth x Length of tunnel

= b x h = 80 x 7 = 560 m².

Hence, area of floor = 560 m².