OP Malhotra Class 9 Maths Solutions Chapter 18 Surface Area and Volume of 3D Solids Ex 18(B)

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 cm², and length 1.26 cm.
Solution:
Area of the cross-section of a rail = 12.8 cm² and length = 1.26 cm
∴ Volume = area × length =12.8 × 1.26 cm³ = 16.128 cm³

Question 2.
Find the area of the cross-section, assuming it to be uniform, of a solid, given that its volume is 92.8 cm³, and length is 6.4 m.
Solution:
Volume of a cross-section of a solid = 92.8 cm³
and its length = 6.4 m = 640 cm
∴ Area of the cross-section = \(\frac{\text { Volume }}{\text { Length }}\)
= \(\frac{92.8}{6.40}\) cm² = 0.145 cm²

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 cm³.

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\(\frac{1}{2}\) × \(\frac{3}{4}\) =\(\frac{7}{2}\) × \(\frac{3}{4}\) = \(\frac{21}{8}\)cm²
∴ Total area = \(\frac{21}{8}\) + \(\frac{3}{2}\) = \(\frac{21+12}{8}\) = \(\frac{33}{8}\) cm²
∴ Volume of the cross-section = Area × Length
= \(\frac{33}{8}\) × 200 = 33 × 25 cm³ = 825 cm³

(ii) Weight of 1 cm³ = 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 cm².

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 cm²

Area of rectangle II = 3.5 × (8 – 3 – 3)
= 3.5 × 2 = 7 cm²
and area of rectangle III = 8 × 1.5 = 12 cm²
∴ Total area of the cross section = 10 + 7 + 12 = 29 cm²

(i) ∴ Volume = Area × Length
= 29 × 250 = 7250 cm³

(ii) Weight of 1 cm³ = 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 cm², 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 cm²
= \(\frac{42}{100 \times 100}\) m²

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 m³
Water in litres = 0.315 × 1000
= 315 litres (1 m² = 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 cm²
∴ Volume = Area × Length
= 20 × 4 = 80 cm³

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 m³
∴ Volume of water in litres = 2250 × 1000 l (1 m³ = 1000 l) = 2250000 litres

Question 8.
The area of cross-section of a pipe is 5.4 cm² 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 cm² 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 m³ = 6.243 m³
∴ Water in litres = 0.243 × 1000 litres (1 m³ = 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 + \(\frac{1}{2}\)πr²
= 16 × 7 + \(\frac{1}{2}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) m² \(\left(r=\frac{7}{2} \mathrm{~m}\right)\)
= 112 + \(\frac{77}{4}\) = \(\frac{448+77}{4}\) = \(\frac{525}{4}\) m²
Depth of water in the tank = 2.4 m
∴ Volume of water = area × depth
= \(\frac{525}{4}\) × 2.4 m³ = 315 m³
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 cm² = 2700 cm²
Length of girder = 5 m
∴ Volume of metal used
= Area × Length
= \(\frac{2700 \times 5}{100 \times 100}\) m³
= \(\frac { 135 }{ 100 }\) m³ = 1.35 m³

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 m³ = 400 m³

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 m³.
(iii) If 40 m³ 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 m²
and area of triangle ABF
= \(\frac { 1 }{ 2 }\) BF × AG
= \(\frac { 1 }{ 2 }\) × 8 × 3 = 12 m²

(i) ∴ Total area of the end-face = 40 + 12 = 52 m²
(ii) Volume of the air space = area × length = 52 × 20 = 1040 m³
(iii) For one worker air is required = 40 m³
∴ No. of workers = \(\frac { 1040 }{ 40 }\) = 26