## 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) cm^{3}

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 cm^{2}

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 cm^{2}

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 = 4m^{2}

**Question 13.**

**Solution:**

Height of the cylindrical bucket (h) = 32cm

Radius (r) = 18cm

Volume of sand filled in it = πr^{2}h

= π x 18 x 18 x 32 cm^{3}

= 10368π cm^{3}

Volume of conical sand = 10368 π cm^{3}

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 = \(\frac { 1 }{ 3 } \) πr^{2}h

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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