Online Education for RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B

Online Education for RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B

These Solutions are part of Online Education RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B.

Other Exercises

Question 1.
Solution:
Radius of the base of a cylinder (r) = 5cm.
and height (h) = 21cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B Q1.1

Question 2.
Solution:
Diameter of the base of the cylinder = 28cm
Radius = \(\frac { 1 }{ 2 } \) x 28 = 14 cm
Height (h) = 40cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B Q2.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B Q2.2

Question 3.
Solution:
Radius of cylinder (r) = 10.5cm
Height (h) = 60cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B Q3.1

Question 4.
Solution:
Diameter of cylinder = 20cm
Radius (r) = \(\frac { 20 }{ 2 } \) = 10cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 04.1

Question 5.
Solution:
Curved surface area of cylinder = 4400 cm²
Circumference of its base = 110 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 05.1

Question 6.
Solution:
The ratio of the radius and height of a cylinder = 2:3
Volume =1617 cm³
Let radius = 2x
and height = 3x.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 06.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 06.2

Question 7.
Solution:
Total surface area of the cylinder = 462 cm²
Curved surface area = \(\frac { 1 }{ 3 } \) x 462 = 154
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 07.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 07.2

Question 8.
Solution:
Total surface area of solid
cylinder = 231 cm²
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 08.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 08.2

Question 9.
Solution:
Sum of radius and height = 37m.
and total surface area = 1628 m²
Let r be the radius
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 09.1

Question 10.
Solution:
Total surface area = 616 cm²
Curved surface area = \(\frac { 616X1 }{ 2 } \) = 308
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 010.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 010.2

Question 11.
Solution:
Volume of gold = 1 cm³
diameter of wire = 0.1 mn
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 011.1

Question 12.
Solution:
Ratio in the radii of two cylinders = 2:3
and ratio in the heights = 5:3
If r1 and r2 and the radii and h1 and h2 are the heights, then
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 012.1

Question 13.
Solution:
Side of square = 12cm
and height = 17.5cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 013.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 013.2

Question 14.
Solution:
Diameter of cylindrical bucket = 28cm
Radius (r) = \(\frac { 28 }{ 8 } \) = 14cm
Height (h) = 72cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 014.1

Question 15.
Solution:
Length of pipe (l) = 1m = 100cm
diameter of pipe = 3cm.
Inner radius = \(\frac { 3 }{ 2 } \) cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 015.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 015.2

Question 16.
Solution:
Internal diameter of cylindrical tube = 10.4 cm
Radius (r) = \(\frac { 10.4 }{ 2 } \) = 5.2cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 016.1

Question 17.
Solution:
Length of barrel (h) = 7cm
Diameter = 5mm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 017.1

Question 18.
Solution:
Diameter of pencil = 7mm
.’. Radius (R) = \(\frac { 7 }{ 2 } \) mm = \(\frac { 7 }{ 20 } \) cm.
and diameter of graphite in it = 1mm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 018.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 018.2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 018.3

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A

Online Education for RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A.

Other Exercises

Question 1.
Solution:
A number in the form of \(\frac { p }{ q }\) where p and q are integers and q ≠ 0, is called a rational number
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 1
are all rational numbers.

Question 2.
Solution:
The given rational number are represented on a number line on given below :
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 2
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 3
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 4

Question 3.
Solution:
We know that, if a and b are two rational numbers, then a rational number between a and b will be
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 5
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 6
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 7

Question 4.
Solution:
Here, n = 3, x = \(\frac { 1 }{ 5 } \), y = \(\frac { 1 }{ 4 } \)
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 8
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 9

Question 5.
Solution:
Here, n=5, x = \(\frac { 2 }{ 5 } \), y = \(\frac { 3 }{ 4 } \)
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 10
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 11
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 12

Question 6.
Solution:
Here, n = 6, x = 3, y = 4
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 13
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 14

Question 7.
Solution:
Here, n = 16, x = 2.1, y = 2.2
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 15
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 16
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 17

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Online Education for RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A

Online Education for RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A.

Other Exercises

Question 1.
Solution:
(i) Length of cuboid (l) = 12cm
Breadth (b) = 8cm
and height (h) = 4.5cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.4

Question 2.
Solution:
Length of closed rectangular cistern (l) = 8m
breadth (b) = 6m
and depth (b) = 2.5m.
(i) .’. Volume of cistern = l.b.h.
= 8 x 6 x 2.5 m³ = 120m³
(ii) Total surface area = 2(lb + bh + hl)
= 2(8 x 6 + 6 x 2.5 + 2.5 x 8) cm²
= 2(48 + 15 + 20)
= 2 x 83 m²
= 166 m² Ans.

Question 3.
Solution:
Length of room (l) = 9m
Breadth (b) = 8m
and height (h) = 6.5m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q3.1

Question 4.
Solution:
Length of pit (l) = 20m
Breadth (b) = 6m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q4.1

Question 5.
Solution:
Length of wall (l) = 8m.
Width (b) = 22.5 cm = \(\frac { 225 }{ 10X100 } =\frac { 9 }{ 40 } m\)
and height (h) = 6m.
Volume of wall = l.b.h.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q5.1

Question 6.
Solution:
Length of wall (l) = 15m.
Width (b) = 30cm = \(\frac { 30 }{ 100 } =\frac { 3 }{ 10 } m\)
Height (h) = 4m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q6.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q6.2

Question 7.
Solution:
Outer length of opened cistern = 1.35m = 135 cm
Breadth = 1.08 m = 108 cm
Depth = 90cm
Thickness of iron = 2.5cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q7.1

Question 8.
Solution:
Depth of river = 2m
width = 45m.
Length of current in 60 minutes = 3km
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q8.1

Question 9.
Solution:
Total cost of box = Rs. 1620
Rate per sq. m = Rs. 30

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q8.2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q8.3

Question 10.
Solution:
Length of room (l) = 10m
Breadth (b) = 10m
Height (h) = 5m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q10.1

Question 11.
Solution:
Length of hall (l) = 20m
Breadth (b) = 16m
and height (h) = 4.5m.
Volume of the air inside the hall
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q11.1

Question 12.
Solution:
Length of class room (l) = 10m
Width (b) = 6.4 m
Height (h) = 5m.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q12.1

Question 13.
Solution:
Volume of cuboid = 1536 m³
Length (l) = 16m
Ratio in breadth and height = 3:2
Let breadth (b) = 3x
their height (h) = 2x
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q13.1

Question 14.
Solution:
Length of cuboid (l) = 14 cm
Breadth (b) = 11 cm .
Let height (h) =x cm
Surface area = 2(lb + bh + hl)
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q14.1

Question 15.
Solution:
(a) Edge of cube (a) = 9m .
(i) volume = a³ = (9)³ m³ = 729 m³
(ii) Lateral surface area = 4a²
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q15.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q15.2

Question 16.
Solution:
Total surface area of a cube = 1176 cm²
Let each edge he ‘a’
then 6a² =1176
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.16.1

Question 17.
Solution:
Lateral surface area of a cube = 900 cm²
Let ‘a’ be the edge of the cube
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q17.1

Question 18.
Solution:
Volume of a cube = 512 cm³
Let ‘a’ be its edge, then
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q18.1

Question 19.
Solution:
Edge of first-cube = 3 cm.
Volume = (3)³ = 27 cm³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q19.1

Question 20.
Solution:
Area of ground = 2 hectares
= 2 x 10000 = 20000 m²
Height of rain falls 5cm = \(\frac { 5 }{ 100 } \)m
∴ Volume of rain water = 20000 x \(\frac { 5 }{ 100 } \) m³
= 1000 m³ Ans.

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.