Interactive ICSE S Chand Maths Class 9 Solutions Chapter 18 Surface Area and Volume of 3D Solids Ex 18(B) engage students in active learning and exploration.

## S Chand Class 9 ICSE Maths Solutions Chapter 18 Surface Area and Volume of 3D Solids Ex 18(B)

Question 1.

Find the volume of a rail of uniform crosssection, given area of cross-section 12.8 cm^{2}, and length 1.26 cm.

Solution:

Area of the cross-section of a rail = 12.8 cm^{2} and length = 1.26 cm

∴ Volume = area × length =12.8 × 1.26 cm^{3} = 16.128 cm^{3}

Question 2.

Find the area of the cross-section, assuming it to be uniform, of a solid, given that its volume is 92.8 cm^{3}, and length is 6.4 m.

Solution:

Volume of a cross-section of a solid = 92.8 cm^{3}

and its length = 6.4 m = 640 cm

∴ Area of the cross-section = \(\frac{\text { Volume }}{\text { Length }}\)

= \(\frac{92.8}{6.40}\) cm^{2} = 0.145 cm^{2}

Question 3.

The dimensions of the L-shaped crosssection of bar, 2 m long, are shown in cm in figure. Find

(i) the volume of the bar.

(ii) the weight, if the material weighs 0.3 kg per cm^{3}.

Solution:

Length of cross-section of bar = 2 m = 200 cm

In the figure produced DC which meets AF at G

Now AG = 4 – \(\frac{1}{2}\) = 3\(\frac{1}{2}\) cm

(i) ∵ Area of the figure of the cross-section = Area of ABCD + Area of GDEF

Area of ABCG = 3\(\frac{1}{2}\) × \(\frac{3}{4}\) =\(\frac{7}{2}\) × \(\frac{3}{4}\) = \(\frac{21}{8}\)cm^{2}

∴ Total area = \(\frac{21}{8}\) + \(\frac{3}{2}\) = \(\frac{21+12}{8}\) = \(\frac{33}{8}\) cm^{2}

∴ Volume of the cross-section = Area × Length

= \(\frac{33}{8}\) × 200 = 33 × 25 cm^{3} = 825 cm^{3}

(ii) Weight of 1 cm^{3} = 0.3 kg

∴ Total weight = 825 × 0.3 kg = 247.5 kg

Question 4.

The dimensions of the cross-section of a girder, 2.5 m long, are shown in cm in the diagram. Find

(i) the volume of the girder,

(ii) the weight, if the material weighs 7.8 gm per cm^{2}.

Solution:

Length of cross section = 2.5 m = 250 cm In the given figure,

On joining LM and JK, we get three rectangles

∴ Area of rectangle I

=4 × 2.5 = 10 cm^{2}

Area of rectangle II = 3.5 × (8 – 3 – 3)

= 3.5 × 2 = 7 cm^{2}

and area of rectangle III = 8 × 1.5 = 12 cm^{2}

∴ Total area of the cross section = 10 + 7 + 12 = 29 cm^{2}

(i) ∴ Volume = Area × Length

= 29 × 250 = 7250 cm^{3}

(ii) Weight of 1 cm^{3} = 7.8 gm

∴ Total weight = 7.8 × 7250 gm = 56550 gm = 56.55 kg = 57 kg (approx)

Question 5.

The cross-area of a pipe is 42 cm^{2}, and water is pouring out of it at the rate of 1.25 m per sec. If the pipe remains full, find the number of litres discharged per minute.

Solution:

Area of cross-section of a pipe = 42 cm^{2}

= \(\frac{42}{100 \times 100}\) m^{2}

Rate of water pouring = 1.25 m per sec

Rate of water per minute = 1.25 m × 60 = 75.00 m per minute

∴ Volume of water in the pipe

= \(\frac{42}{100 \times 100}\) × 75 = 0.315 m^{3}

Water in litres = 0.315 × 1000

= 315 litres (1 m^{2} = 1000 l)

Question 6.

The figure shows a solid of uniform crosssection. Find the volume of the solid. All measurements are in centimetres. Assume that all angles in the figure are right angles.

Solution:

Length of the cross-section = 4 c

Area of the cross-section = 4 × 2 + 6 × 2

= 8 + 12 = 20 cm^{2}

∴ Volume = Area × Length

= 20 × 4 = 80 cm^{3}

Question 7.

A swimming pool is 50 m long and 15 m wide. Its shallow and deep ends are 1\(\frac{1}{2}\) m and 4\(\frac{21}{8}\) m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.

Solution:

Length of swimming pool (l) = 50 m and width (b) = 15 m

Depth of water in one side = 1\(\frac{1}{2}\) m

and other side = 4\(\frac{1}{2}\) m

∴ Mean depth (h) = \(\frac{1}{2}\) \(\left[1 \frac{1}{2}+4 \frac{1}{2}\right]\)m

= \(\frac{1}{2}\) × 6 = 3 m

∴ Volume of the pool = lbh

= 50 + 15 × 3 = 2250 m^{3}

∴ Volume of water in litres = 2250 × 1000 l (1 m^{3} = 1000 l) = 2250000 litres

Question 8.

The area of cross-section of a pipe is 5.4 cm^{2} and water is pumped out of it the rate of 27 km / h. Find in litres the volume of water which flows out of the pipe in one minute.

Solution:

Area of the cross section = 5.4 cm^{2} Flows of water = 27 km / h

Flow of water in 1 minute = \(\frac{27 \times 1000}{60}\)m = 450 m

∴ Volume of the water = \(\frac{5.4}{100 \times 100}\) × 450 m^{3} = 6.243 m^{3}

∴ Water in litres = 0.243 × 1000 litres (1 m^{3} = 1000 l) = 243 litres

Question 9.

The horizontal cross-section of a water tank is in the shape of a rectangle with a semicircle at one end, as shown in figure. The water is 2.4 m deep in the tank. Calculate the volume of water in the tank in gallons.

( Take π to be \(\frac{22}{7}\) and 1 gallon to be 4.5 litres)

Solution:

Area of the cross-section of the tank = area of rectangle ABCD + area of semicircle or BC

= l × b + \(\frac{1}{2}\)πr^{2}

= 16 × 7 + \(\frac{1}{2}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) m^{2} \(\left(r=\frac{7}{2} \mathrm{~m}\right)\)

= 112 + \(\frac{77}{4}\) = \(\frac{448+77}{4}\) = \(\frac{525}{4}\) m^{2}

Depth of water in the tank = 2.4 m

∴ Volume of water = area × depth

= \(\frac{525}{4}\) × 2.4 m^{3} = 315 m^{3}

Water in gallons = 315 × 1000 l

=\(\frac{315 \times 1000}{4.5}\) gallons

& =70000 gallon

Question 10.

In the figure, shows the end face of a triangular girder which is 5 m long. Find the volume of metal used in casting it. (Units in cm).

Solution:

Cross-section of the girder is triangular whose base (b) = 60 m and height (h) = 90 cm

∴ Area of the face of the girder

= \(\frac{1}{2}\) × b × h = \(\frac{1}{2}\) × 60 × 90 cm^{2} = 2700 cm^{2}

Length of girder = 5 m

∴ Volume of metal used

= Area × Length

= \(\frac{2700 \times 5}{100 \times 100}\) m^{3}

= \(\frac { 135 }{ 100 }\) m^{3} = 1.35 m^{3}

Question 11.

In the figure shows the end-view of a swimming pool which is 10 m wide. Calculate the volume of water when it is full of water \(\frac { 1 }{ 2 }\) m from the top.

Solution:

Width of pool = 10 m

Water = \(\frac { 1 }{ 2 }\) m from the top

Length = 20 m

∴ Height of water from one side = 2 – \(\frac { 1 }{ 2 }\)

= 1\(\frac { 1 }{ 2 }\) m

∴ Average height = \(\frac { 1 }{ 2 }\) \(\left(1 \frac{1}{2}+2 \frac{1}{2}\right)\)

= \(\frac { 1 }{ 2 }\) × 4 = 2 m

∴ Volume of water in the pool = l × b × h

= 20 × 10 × 2 m^{3} = 400 m^{3}

Question 12.

In the figure shows the end-face of a garage which is 20 m long.

(i) Find the area of the end-face.

(ii) Find the volume of air-space of the garage in m^{3}.

(iii) If 40 m^{3} of air is required per worker, how many workers can be employed in this garage ?

Solution:

Length of garage = 20 m

By joining BF, we get the end face of the garage, one rectangle and one triangle

∴ Area of rectangle BCEF = CE × BC = 8 × 5 = 40 m^{2}

and area of triangle ABF

= \(\frac { 1 }{ 2 }\) BF × AG

= \(\frac { 1 }{ 2 }\) × 8 × 3 = 12 m^{2}

(i) ∴ Total area of the end-face = 40 + 12 = 52 m^{2}

(ii) Volume of the air space = area × length = 52 × 20 = 1040 m^{3}

(iii) For one worker air is required = 40 m^{3}

∴ No. of workers = \(\frac { 1040 }{ 40 }\) = 26