RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.1 Solutions: Wherever we look, usually we see solids. So far, in all our studies, we have been dealing with figures that can be easily drawn on our notebooks or blackboards. These are called plane figures. We have understood what rectangles, squares, and circles are, what we mean by their perimeters and areas, and how we can find them. We have learned these in earlier classes. It would be interesting to see what happens if we cut out many of these plane figures of the same shape and size from cardboard sheets and stack them up in a vertical pile. Through this process, we shall obtain some solid figures (briefly called solids) such as a cuboid, a cylinder, etc. Know more about this chapter here.
Download RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.1 Solutions
Important Definition for RS Aggarwal Chapter 15 Class 9 Maths Ex 15a Solutions
|
Name of the Solid Figure |
Formulas |
|
Cuboid |
LSA: 2h(l + b) |
|
Cube |
LSA: 4a2 |
|
Right Circular Cylinder |
LSA: 2(π × r × h) |
|
Right Pyramid |
LSA: ½ × p × l |
|
Prism |
LSA: p × h |
|
Right Circular Cone |
LSA: πrl |
|
Hemisphere |
LSA: 2 × π × r2 |
|
Sphere |
LSA: 4 × π × r2 |
Access The RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.1 Solutions
(1) Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:
(i) Length = 12 cm, breadth = 8 cm and height = 4.5 cm
Volume of cuboid = (lbh) cm3
=12×8×4.5 cm3
=432 cm3
Lateral surface area =[2(l+b)×h] cm2
=[2(12+8)×4.5]
=40×4.5
=180 cm2
Total surface area =2(lb+bh+hl) cm2
=2{12×8+8×4.5+4.5×12} cm2
=2(96+36+54) cm2
=372 cm2
Hence volume of cuboid is 432 cm3, lateral surface area is 180 cm2 and total surface area is 372 cm2
(ii) Length = 26 m, breadth = 14 m and height = 6.5m
Volume of cuboid = (lbh) m3
=26×14×6.5 m3
=2366 m3
Lateral surface area =[2(l+b)×h] m2
=[2(26+14)×6.5]
=80×6.5
=520 m2
Total surface area =2(lb+bh+hl) m2
= 2{26×14+14×6.5+6.5×26} m2
= 2(364+91+169) m2
= 1248 m2
Hence volume of cuboid is 2366 m3, lateral surface area is 520 m2 and total surface area is 1248 m2
(iii) Length = 15 m, breadth = 6 m and height = 5 dm
5 dm = 50 cm = 0.5 m
Volume of cuboid = (lbh) m3
=15×6×0.50 cm3
=45 m3
Lateral surface area =[2(l+b)×h] m2
=[2(15+6)×0.50]
=42×0.50
=21 cm2
Total surface area =2(lb+bh+hl) m2
= 2{15×6+6×0.50+0.50×15} m2
= 2(90+3+7.5) m2
= 201 m2
Hence the volume of the cuboid is 45 m3, the lateral surface area is 21 m2 and the total surface area is 201m2
(iv) Length = 24m, breadth = 25 cm and height = 6m
breadth = 25 cm = 0.25 m
Volume of cuboid = (lbh) m3
=24×0.25×6 m3
=36 m3
Lateral surface area =[2(l+b)×h] m2
=[2(24+0.25)×6]
=48.5×6
=291 m2
Total surface area =2(lb+bh+hl) m2
= 2{24×0.25+0.25×6+6×24} m2
= 2(6+1.5+144) m2
=303 m2
(2) A match box measures 4 cm×2.5 cm×1.5 cm. What is the volume of a packet containing 12 such matchbox?
Volume of match box = 4×2.5 ×1.5=15 cm3
Volume of packet containing 12 match box = 15×12=180 cm3
(3) A cuboid water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold?
Volume of tank = 6×5 ×4.5
=135 m3
Water in litre = 135×1000
=135000 litres [∵ 1 m3 =1000 litres]
(4) The capacity of the Cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10m and 2.5 m. (Given 1000 litres = 1m3)
Let the breadth of the tank is x, then
Volume of tank =10×x×2.5
=25x m3
Water in litre =25x×1000
=25000x litres
According to the question,
⇒ 25000x=50000
⇒ x = 2 m
(5) A godown measures 40m×25m×15m . Find the maximum number of wooden crates, each measuring1.5m×1.25m×0.5m, that can be stored in the godown.
Volume of godown =40×25×15
=15000 m3
The volume of one crate = 1.5×1.25×0.5
= 0.9375 m3
Number of crates =15000/0.9375
= 16000
Hence 16000 crates can be stored in the godown.
(6) How many planks of dimension (5m×25cm×10cm) can be stored in a pit which is 20m long, 6m wide and 80 cm deep?
Volume of pit = 20 m×6 m×80 cm
=2000 cm×600 cm×80 cm
=96000000 cm3
The volume of one plank = 5m×25cm×10cm
=500cm×25cm×10cm
=125000 cm3
Number of planks = 96000000/125000
= 768
Hence total 768 planks can be stored.
(7) How many bricks will be required to construct a wall 8 m long, 6 m high, and 22.5 cm thick if each brick measures(25cm×11.25cm×6cm)?
Volume of wall = 8m × 6m × 22.5cm
= 800 cm×600 cm×22.5 cm
=10800000 cm3
Volume of one brick = 25cm × 11.25cm × 6cm
=1687.5 cm3
Number of bricks = 6400
Hence, 6400 bricks are required.
(8) Find the capacity of a closed rectangular cistern whose length is 8m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.
Volume of cistern = 8 m×6 m×2.5 m
=120 m3
Total surface area =2(lb+bh+hl)
=2(8×6+6×2.5+2.5×8)
=2(48+15+20)
=2(83)
=166 m2
Hence, Volume is 120 m3 and 166 m2 iron sheet is required.
(9) The dimensions of the room are(9m×8m×6.5m). It has one door of dimensions (2m×1.5m) and two windows, each of dimensions(1.5m×1m). Find the cost of white washing the walls at Rs. 25 per square metre.
Area of room’s wall =[2(l+b)×h]
=[2(9m+8m)×6.5m]
=221 m2
And, Area of one door + two windows
= (2m×1.5m)+2(1.5m×1m)
=(3+3)m2
Net area for white washing = 221-6
=215 m2
Cost of white washing = 215×25
= 5375
Hence the cost of white washing is Rs. 5375
(10) A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring(22 cm×12.5 cm×7.5 cm) . If 1/12 of the total volume of wall consists of mortar, now many bricks are there in the wall?
Volume of wall = (15 m×30 cm×4 m)
=(1500 cm×30 cm×400 cm)
=18000000 cm3
Volume of mortar=1/12×18000000
=1500000 cm3
Volume of wall without mortar = (18000000-1500000) cm3
=16500000 cm3
Volume of one brick = (22 cm×12.5 cm×7.5 cm)
=2062.5 cm3
Number of bricks = 16500000/2062.5
= 8000
Hence 8000 bricks are in the wall.
(11) How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm3 of iron weight 15 g, find the weight of the empty box in kilograms.
External length = 36 cm
External breadth = 25 cm
External height = 16.5 cm
Internal length = 36 – 2(1.5) = 33 cm
Internal breadth = 25 – 2(1.5) = 22 cm
Internal height = 16.5 – 1.5 = 15 cm
Internal volume of the box = (33×22×15)
=10890 cm3
Volume of iron sheet = 14850-10890
=3960 cm3
Weight of sheet = 3960×15g
= 59.4 kg
Hence 3960 cm3 iron sheet is required and weight of sheet is 59.4 kg
(12) A box made of sheet metal costs Rs. 6480 at Rs. 120 per square metre. If the box is 5 m long and 3 m wide, find its height.
Let height = x metre
Total surface area of box =2(lb+bh+hl)
=2(5×3+3×x+x×5)=2(15+8x)
Cost of metal at Rs. 120 =2(15+8x)×120
=3600+1920x
According to question, =3600+1920x = 6480
1920x =6480-3600 =2880
x=2880/1920 = 1.5 m
Hence height of box is 1.5 m
(13) The volume of cuboid is 1536 m3 . Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.
Length = 16 m, breadth = 3x and height = 2x
Volume of cuboid = lbh m3
=16×3x×2x
= 96 x2m3
By question,=96 x2m3
=1536 m3
⇒ x2 =1536/96 = 16
⇒ x = 4
Therefore breadth =3×4=12 m and height =2×4=8 m
(14) How many persons can be accommodated in dining hall of dimensions (20 m×16 m×4.5 m), assuming that each person requires 5 cubic metres of air?
Volume of dining hall = 20 m×16 m×4.5 m
=1440 m3
Each person required 5 cubic metre air, therefore
Number of person in hall = 1440/5 = 288
Hence, 288 person can accommodate in dining hall.
(15) A class room is 10 m long, 6.4 m wide and 5 m high. If each student be given 1.6 m2 of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?
Area of the class = 10 m×6.4 m
=64 m2
Each student needs 1.6 m2 area, therefore
Number of student in class = 64/1.6 = 40
Air for each student = Floor area×height
= 1.6×5 = 8 m3
Hence, 40 Student can accommodate in class and each student get 8m3 air.
(16) The surface area of cuboid is 758 cm2. Its length and breadth are 14 cm and 11 cm respectively. Find its height.
Let height = x cm
Surface are of cuboid =2(lb+bh+hl)
=2(14×11+11×x+x×14)
=2(154+25x)
=(308+50x) cm2
According to question, 308+50x =758
50x =758-308
x = 450
x = 9 cm
Hence height of cuboid is 9 cm.
(17) In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.
1 hectare = 10000 m2
Area of ground = 2 hectare = 20000 m2
Volume of water = 20000 m2 ×5 cm
= 20000 m2 ×0.05 m
=1000 m3
Hence, 1000 m3 of rain falls.
(18) Find the volume, the lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures 9 m. (Take √3=1.73)
Volume of cube = (side)3 = 93
=729 m3
Total surface area = 6 ×(side)3
= 6×92
= 6×81
= 486 m2
Diagonal = √3× (side)
= 1.73×9
= 15.57 m
Hence volume of cube is 729 m3 ,total surface area is 486 m2 and diagonal is 15.57m.
(19) The total surface area of a cube is 1176 cm2 . Find its volume.
Let side is x cm.
Total surface area of cube = 6 ×(x)3 = 1176
(x)2 = 1176/6 = 196 ⇒ x = 14
Volume = (14)3 = 2744 cm3
Hence volume is 2744 cm3.
(20) The lateral surface area of a cube is 900 cm2 . Find its volume.
Let side of cube is x cm.
Lateral surface area =[2(l+b)×h] cm2
=[2(x+x)×x]=900
= 4x2 = 900
= x2 = 225
⇒ x = 15
Volume = (15)3 = 3375 cm3
(21) The volume of a cube is 512 cm3 . Find its surface area.
Let side of cube is x cm
Volume = (x)3
= 512 cm3
⇒ x = 8
Surface area = 6× (x)2
= 6×64
⇒ 384 cm2
Hence surface area is 384 cm2 .
(22) Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.
Volume of first cube =33
= 27 cm3
Volume of second cube = 43
= 64 cm3
Volume of third cube =53
= 125 cm3
Total volume = 27 + 64 + 125
= 216
Volume of large cube = (x)3 = 216
x = 6 cm
Lateral surface area = [2(x+x)×x]
= 4x2
= 4×36
= 144 cm2
Hence lateral surface area of the new cube is 144 cm2 .
(27) Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hour. How much area will it irrigate in 30 minute, if 9 cm of standing water is desired?
Distance covered by water in 30 minute
=speed of water×time
=20×1/2 h
=10km
Volume of water which flown in 30 min
=30 dm ×12 dm×10km
=3 m×1.2 m×10000 m
=36000 m3
Let irrigation area is x m2 , then
x m2 ×9 cm = 36000 m3
x m2 ×0.09 m = 36000 m3
x = 400000
Hence irrigation area is 400000 m2
(28) A solid metallic cuboid of dimensions (9 m×8 m×2 m) is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.
Volume of cuboid = (9 m×8 m×2 m)
=144 m3
Volume of cube which side is 2 m
= 23
=8m3
Number of cubes = 144/8 =18
Hence 18 cubes formed.
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