The document Ex 13.7 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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**Note: **Assume Ï€ = (22/7), unless stated otherwise.

**Question 1. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm. Solution:** (i) Here, radius of the cone r = 6 cm Height (h) = 7 cm

(ii) Here, radius of the cone (r) = 3.5 cm = (35/10) cm

Height (h) = 12 cm

âˆ´ Volume of the cone

= 11 x 7 x 2 cm^{3} = 154 cm^{3}

**Question 2. Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm Solution:** (i) Here, r = 7 and l = 25 cm

âˆ´ Volume of the conical vessel

Thus, the required capacity of the conical vessel is 1.232 l.

(ii) Here, height (h) = 12 cm and l = 13 cm

Now, volume of the conical vessel

âˆ´ Capacity of the conical vessel =

Thus, the required capacity of the conical vessel is

**Question 3. The height of a cone is 15 cm. If its volume is 1570 cm ^{3}, find the radius of the base. (Use Ï€ = 3.14) Solution: **Here, height of the cone (h) = 15 cm

Volume of the cone (v) = 1570 cm

Let the radius of the base be â€˜râ€™ cm.

â‡’ r^{2} = 10^{2}

â‡’ r = 10 cm

Thus, the required radius of the base is 10 cm.

**Question 4. If the volume of a right circular cone of height 9 cm is 48p cm ^{3}, find the diameter of its base. Solution:** Volume of the cone = 48 p cm

Height of the cone (h) = 9 cm

Let â€˜râ€™ be its base radius.

âˆµ Diameter = 2 x Radius

âˆ´ Diameter of the base of the cone = 2 x 4 = 8 cm

**Question 5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? Solution:** Here, diameter of the conical pit = 3.5 m

= 38.5 m^{3}

âˆ´ 1000 cm^{3} = 1 l and 1000000 cm^{3} = 1m^{3}

âˆ´ 1000 x 1000 cm^{3} = 1000 l = 1 kl

Also 1000 x 1000 cm^{3 }= 1 m^{3}

â‡’ 1 m^{3} = 1 kl

â‡’ 38.5 m^{3} = 38.5 kl

Thus, the capacity of the conical pit is 38.5 kl.

**Question 6. The volume of right circular cone is 9856 cm ^{3}. If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone. Solution: **Volume of the cone (v) = 9856 cm

Diameter of the base = 28 cm

â‡’ Radius of the base =(28/2)cm = 14 cm

(i) To find the height Let the height of the cone be â€˜hâ€™ cm.

= 16 x 3 cm = 48 cm Thus, the required height is 48 cm.

(ii) To find the slant height Let the slant height be â€˜â„“â€™ cm.

âˆµ (Slant height)^{2} = (Radius)^{2} + (Height)^{2}

âˆ´ â„“^{2} = 14^{2} + 48^{2} = 196 + 2304 = 2500 = (50)^{2}

â‡’ â„“ = 50

Thus, the required height = 50 cm.

(iii) To find the curved surface area

âˆµ The curved surface area of a cone is given by Ï€rl

âˆ´ Curved surface area = (22/7) x 14 x 50 cm^{2}

= 22 x 2 x 50 cm^{2 }

= 2200 cm^{2 }

Thus, the curved surface area of the cone is 2200 cm^{2}.

**Question 7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained. Solution: **Sides of the right triangle are 5 cm, 12 cm and 13 cm.

âˆµ The right angled triangle is revolved about the 12 cm side.

âˆ´ Its height is 12 cm and base is 5 cm.

Thus, we have Radius of the base of the cone so formed (r) = 5 cm

Height (h) = 12 cm

Slant height = 13 cm

âˆ´ Volume of the cone so obtained = 1/3Ï€r

= Ï€ x 100 cm^{3 } [âˆµ Ï€ = 22/7]

= 100p cm^{3}

Thus, the required volume of the cone is 100p cm^{3}.

**Question 8. If the triangle ABC in the question 7 a bove is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8. Solution: **Since the right triangle is revolved about the side 5 cm.

âˆ´ Height of the cone so obtained (h) = 5 cm

Radius of the cone (r) = 12 cm

âˆ´ Volume

Thus, the required ratio is 5 : 12.

**Question 9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to Ï€rotect it from rain. Find the area of the canvas required. Solution: **Here the heap of wheat is in the form of a cone such that Base diameter = 10.5 m

â‡’ Base radius (r) = (10.5/2)m = (105/20) m

Height (h) = 3 m Volume of the heap

Thus, the required volume = 86.625 m^{3 }

Area of the canvas

âˆµ The area of the canvas to cover the heap must be equal to the curved surface area of the conical heap.

= 6.05 m (apÏ€rox.)

Now, = 11 x 1.5 x 6.05 m^{2}

= 99.825 m^{2}

Thus, the required area of the canvas is 99.825 m^{2}.

**VOLUME OF A SPHERE**

**REMEMBER**

(i) Volume of a sphere = (4/3) Ï€r^{3}, where â€˜râ€™ is the radius of the sphere.

(ii) Volume of a hemisphere = (2/3)Ï€r^{3}, where â€˜râ€™ is the radius of the hemisphere.

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