RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

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 13C.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13A
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13B
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13C
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13D

Question 1.
Solution:
Radius of base (r) = 35cm
and height (h) = 84cm.

Question 2.
Solution:
Height of cone (h) = 6cm
Slant height (l) = 10cm.

Question 3.
Solution:
Volume of right circular cone = (100 π) cm³
Height (h) = 12cm.
Let r be the radius of the cone

Question 4.
Solution:
Circumference of the base = 44cm

Question 5.
Solution:
Slant height of the cone (l) = 25cm
Curved surface area = 550 cm²
Let r be the radius
πrl = curved surface area

Question 6.
Solution:
Radius.of base (r) = 35cm.
Slant height (l) = 37cm.
We know that

Question 7.
Solution:
Curved surface area = 4070 cm²
Diameter of the base = 70cm

Question 8.
Solution:
Radius of the conical tent = 7m
and height = 24 m.

Question 9.
Solution:
Radius of the first cone (r) = 1.6 cm.
and height (h) = 3.6 cm.

Question 10.
Solution:
Ratio in their heights =1:3
and ratio in their radii = 3:1
Let h1,h2 he their height and r1,r2 be their radii, then

The ratio between their volumes is 3:1
hence proved

Question 11.
Solution:
Diameter of the tent = 105m

Question 12.
Solution:
No. of persons to be s accommodated =11
Area to be required for each person = 4m²

Question 13.
Solution:
Height of the cylindrical bucket (h) = 32cm
Radius (r) = 18cm
Volume of sand filled in it = πr²h
= π x 18 x 18 x 32 cm³
= 10368π cm³
Volume of conical sand = 10368 π cm³
Height of cone = 24 cm

Question 14.
Solution:
Let h be the height and r be the radius of the cylinder and cone.
Curved surface area of cylinder = 2πrh
and curved surface area of cone = πrl

Question 15.
Solution:
Diameter of the pillar = 20cm
Radius (r) = \(\frac { 20 }{ 2 } \) = 10cm

Question 16.
Solution:
Height of the bigger cone (H) = 30cm
By cutting a small cone from it, then volume of smaller cone = \(\frac { 1 }{ 27 } \) of volume of big cone

Let radius and height of the smaller cone be r and h
and radius and height of the bigger cone be R and H.


Hence at the height of 20cm from the base it was cut off. Ans.

Question 17.
Solution:
Height of the cylinder (h) = 10cm.
Radius (r) = 6cm.
Height of the cone = 10cm

Question 18.
Solution:
Diameter of conical vessel = 40cm
Radius (r) = \(\frac { 40 }{ 2 } \) = 20cm
and depth (h) = 24cm.
.’. Volume = \(\frac { 1 }{ 3 } \) πr²h

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

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