Right Circular Cone, Sphere - Surface Area and Volumes , Class 10, Mathematics | Extra Documents, Videos & Tests for Class 10 PDF Download

RIGHT CIRCULAR CONE

Solids like an ice-cream cone, a conical tent, a conical vessel, a clown's cap etc. are said to be in conical shape. In mathematical terms, a right circular cone is a solid generated by revolving a right-angled triangle about one of the sides containing the right angle.
Let a triangle AOC revolve about it's side OC, so as to describe a right circular cone, as shown in the figure.

Cones Not Right Circular

There are two cases when we cannot call a cone a right circular cone.

Case-I : The figure shown below is not a right circular cone because the line joining its vertex to the centre of its base is not at right angle to the base.
Base
 

Case-II : The figure shown below is not a right circular cone because the base is not circular.

REMARK : Unless stated otherwise, by 'cone' in this chapter, we shall mean 'a right circular cone'.

The following are definitions of some terms related to right circular cone :

(i) The fixed point O is called the vertex of the cone.

(ii) The fixed line OC is called the axis of the cone.

(iii) A right circular cone has a plane end, which is in circular shape. This is called the base of the cone. The vertex of a right circular cone is farthest from its base.

(iv) The length of the line segment joining the vertex to the centre of the base is called the height of the cone. Length OC is the height of the cone.

(v) The length of the line segment joining the vertex to any point on the circular edge of the base, is called the slant height of the cone.
Length OA is slant height of the cone.

(vi) The radius AC of the base circle is called the radius of the cone.

Relation Between Slant Height, Radius and Vertical Height.

Let us take a right circular cone with vertex at O, vertical height h, slant height l and radius r. A is any point on the rim of the base of the cone and C is the centre of the base. Here, OC = h, AC = r and OA = l.

The cone is right circular and therefore, OC is at right l angle to the base of the cone. So, we have OC ^ CA, i.e., DOCA is right angled at C.
Then by Pythagoras theorem, we have  : l² = r² + h² 

Formulae For a right circular cone of Radius = r, Height = h & Slant Height = l, we have :

Area of the curved (lateral) surface  

Total Surface Area of cone = (Corved surface Area + Area of Base)

Hollow Right Circular Cone Suppose a sector of a circle is folded to make the radii coincide, then we get a hollow right circular cone. In such a cone;

(i) Centre of the circle is vertex of the cone.
(ii) Radius of the circle is slant height of the cone.
(iii) Length of arc AB is the circumference of the base of the cone.
(iv) Area of the sector is the curved surface area of the cone.

Ex.11 The total surface area of a right circular cone of slant height 13 cm is 90p cm².

Calculate : (i) its radius in cm, (ii) its volume in cm3, in terms of p.

Sol. Given : slant height, l = 13 cm.

Let, radius = r cm and height = h cm. (i) Total surface area =

 ⇒ r² + 13r – 90 = 0 ⇒ (r + 18) (r – 5) = 0 ⇒ r = 5  [Neglecting r = – 18, as radius cannot be negative]

∴ Radius of the cone =

(ii)
=

∴ Volume of the cone = 

= 100π cm³.

Ex.12 A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of the sand. If the height of the conical heap is 24 cm, find : (i) its radius, (ii) its slant height.

Sol. Height of cylindrical bucket, H = 32 cm.
Radius of cylindrical bucket, R = 18 cm.

Volume of sand = 

(i) Height of conical heap, h = 24 cm.
Let the radius of the conical heap be r cm.
Then, volume of conical heap = 

Now, Volume of conical heap = Volume of sand

 = (18 × 18 × 4)
 = (18 × 2) cm = 36 cm.

∴ Radius of the heap =36 cm.
(ii) Slant height, 

Ex.14 The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be  of the volume of the given cone, at what height, above the base is the section cut?

Sol. Let OAB be the given cone of height, H = 30 cm and base radius R cm. Let this cone be cut by the plane CND to obtain the cone OCD with height h cm and base radius r cm.
Then, DOND ~ DOMB.
So,

Volume of cone OCD = 1
27 ×  Volume of cone OAB

   [From (i)]

⇒ 9h³ = 9000 ⇒ h³ = 1000 ⇒ h = 10.

∴ Height of the cone OCD = 10 cm.
Hence, the section is cut at the height of (30 – 10) cm, i.e., 20 cm from the base.

Ex.15 From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.

Sol. Radius, r = 7 cm.
Height of the cylinder, H = 30 cm.
Height of the cone, h = 24 cm.
Slant height of the cone,  

(i) Volume of the remaining solid =  (Volume of the cylinder) – (Volume of the cone)
Surface Area and Volumes – Aadhar

= (22 × 7 × 22) cm3 = 3388 cm3.

(ii) Total surface area of the remaining solid = Curved surface area of cylinder + Curved surface area of cone + Area of (upper) circular base of cylinder


= (22 ×  92) cm² = 2024 cm².

SPHERE

Objects like football, volleyball, throw-ball etc. are said to have the shape of a sphere.
In mathematical terms, a sphere is a solid generated by revolving a circle about any of its diameters.
Let a thin circular disc of card board with centre O and radius r revolve about its diameter AOB to describe a sphere as shown in figure.

  
Here, O is called the centre of the sphere and r is radius of the sphere. Also, the line segment AB is a diameter of the sphere.
Formulae For a solid sphere of radius = r, we have :

SPHERICAL SHELL

The solid enclosed between two concentric spheres is called a spherical shell.

Formulae For a spherical shell with external radius = R and internal radius = r, we have : 

Thickness of shell = (R- r) units.
Outer surface area = 4πR² sq. units,
Inner surface area = 4πr² sq. units,
Volume of material = 4/3 π(R³ - r³) cubic units.

HEMISPHERE

When a plane through the centre of a sphere cuts it into two equal parts, then each part is called a hemisphere.
From a solid sphere, the obtained hemisphere is also a solid and it has a base as shown in fig.

Formulae For a hemisphere of radius r, we have :

Curved surface area = 2πr² sq. units.
Total Surface Area = (2πr² + πr²) = 3πr² sq. units.

Volume = 2/3 πr³ cubic units.

HEMISPHERICAL SHELL

The solid enclosed between two concentric hemispheres is called a hemispherical shell.

Formulae For a hemispherical shell of external radius = R and internal radius = r, we have :

Thickness of the shell = (R - r) units.
Outer curved surface area = (2πR²) sq. units.
Inner curved surface area = (2πr²) sq. units.
Total surface area = 2πR² + 2πr² + π(R² - r²) = π(3R² + r²) sq. units.

Volume of the material = 2/3 π(R³ - r³)cubic units.

Ex.16 A solid consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of the cone is 4 cm. Give your answer to the nearest cm³ (Take p = 22/7)

Sol. Radius of the cylinder = 3 cm and its height = 6 cm.
Volume of water in the cylinder, when full =
Volume of solid consisting of cone and hemisphere = (Volume of hemi-sphere) + (Volume of cone)

Volume of water displaced from cylinder = Volume of solid consisting of cone and hemisphere

Volume of water left in the cylinder after placing the solid into it

= 136.19 cm³.
Hence, the volume of water left in the cylinder to the nearest cm³ is 136 cm³.

Ex.17 The given figure shows the cross-section of an ice-cream cone consisting of a cone surmounted by a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 10.5 cm. The outer shell ABCDFE is shaded and is not filled with ice-cream. AE = DC = 0.5 cm, AB||EF and BC||FD. Calculate:
 (i) the volume of the ice-cream in the cone (the unshaded portion including the hemisphere) in cm³;
 (ii) the volume of the outer shell (the shaded portion) in cm³. Give your answer to the nearest cm³.

Sol. Radius of hemisphere, R = AG = 3.5 cm.
External radius of conical shell, R = AG = 3.5 cm.
Internal radius of conical shell, r = EG = (AG – AE) = (3.5 – 0.5) cm = 3 cm External height of conical shell, H = BG = 10.5 cm.
Now, ΔABG ~ ΔEFG.
∴  FG = 9 cm.

So, internal height of conical shell, h = FG = 9 cm. (i) Volume of ice-cream = Volume of hemisphere + Internal volume of conical shell

 = 
= 

= 

= 174.69 cm³ = 175 cm³. (to the nearest cm³)

(ii) Volume of the shell = External volume – Internal volume

= 
= 

= 


= 49.89 cm³
= 50 cm³  (to the nearest cm³)

Ex.19 The outer and inner diameters of a hemispherical bowl are 17 cm and 15 cm respectively. Find the cost of polishing it all over at 25 paise per cm². ( Take π = 22/7). 
Sol. Outer radius =
 cm, Inner radius =

Area of outer surface = 

Area of inner surface =

Area of the ring at the top

∴ Total area to be polished 

 

= 858 cm².

∴ Cost of polishing the bowl = Rs

 = Rs. 214.50.

FRUSTUM

FRUSTRUM OF A RIGHT CIRCULAR CONE

In our day-to-day life we come across a number of solids of the shape as shown in the figure. For example,a bucket or a glass tumbler. We observe that this type of solid is a part of a right circular cone and is obtained when the cone is cut by, a plane parallel to the base of the cone.

If a right circular cone is cut off by a plane parallel to its base, the portion of the cone between the plane and the base of the cone is called a frustum of the cone.
We can see this process from the figures given below: The lower portion in figure is the frustum of the cone. It has two parallel flat circular bases, mark as Base (1) and Base (2). A curved surface joins the two bases.

The line segment MN joining the centres of the two bases is called the height of the frustum. Diameter CD of Base (2) is parallel to diameter AB of base

(1). Each of the line segments AC and BD is called the slant height of the frustum. We observe from the figures (i) and (ii) that,
1. Height of the frustum = (the height of the cone OAB) – (the height of the cone OCD)
2. Slant height of the frustum = (the slant height of the cone OAB) - (the slant height of the cone OCD) 8 Volume of a Frustum of a Right Circular Cone Let h be the height ; r₁ and r₂ be the radii of the two bases (r₁ > r₂₎ of frustrum of a right circular cone.
The frustum is made from the complete cone OAB by cutting off the conical part OCD. Let h₁ be the height of the cone OAB and h₂ be the height of the cone OCD.
Here, h₂ = h₁ – h.
Since right angled triangles OND and OMB are similar, therefore, we have:

Volume V of the frustum of cone = Volume of the cone OAB – Volume of the cone OCD

 = 

= 
 ∴

Note : Volume V =

=   { (area of base 1) + (area of base 2) }

• Curved Surface Area of a Frustum of a Right Circular Cone

 (i)

Let h be the height, l be the slant height and r₁, r₂ be the radii of the bases where r₁ > r₂.

In figure (i), we observe EB = r₁ – r₂
and
∴

In figure (ii), we have OAB as the complete cone from which cone OCD is cut off to make the frustum ABDC.

Let l1 be the slant height of the cone OAB and l2 be the slant height of the cone OCD.

Since, DOND and DOMB are similar,

 
⇒
⇒
⇒
⇒

Curved surface area of frustum ABDC = (Curved surface area of cone OAB) – (Curved surface area of cone OCD)

Therefore, curved surface area of frustum =

Total surface Area of a Frustum of a solid Right Circular Cone

Let h be the height, l be the slant height and r₁, r₂ the radii of the bases where r₁ > r₂ as shown in figure.

Total surface area of this frustum = Curved surface area + Area of Base 1 + Area of Base 2

Area of the Metal Sheet Used To Make a Bucket A bucket is in the shape of a frustrum of a right circular hollow cone.

Let h be the depth, l be the slant height, r₁ be the radius of the top and r₂ be the radius of the bottom as shown in figure

The area of the metal sheet used for making the bucket = Outer (or inner) curved surface area + Area of bottom
=

Ex. 1. A bucket is in the form of a frustum of a cone, its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm respectively.
 Find how many litres of water can the bucket hold ? (Take π = 22/7)

Sol. 
R = 28 cm
r = 21 cm
h = 15 cm
Capacity of the bucket = 

=  × 22 × 15 × {(28)2 + (21)2 + (28) (21)} cm³ 
= × 5 × {784 + 441 + 588} cm³
= × 5 × 1813 cm3= 22 × 5 × 259 cm³
= 28490 cm³ 

= 28·49 litres

Ex. 2.  A container made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container at the rate of Rs. 15 per litre and the cost of the metal sheet used, if it costs Rs. 5 per 100 cm². (Take p 3·14)

Sol. 
R = 20 cm,
r = 8cm,
h = 16 cm

Volume of contamer = 1/3 ×  (3.14) × 16 {400 + 64 + 160} cm³

= 3.14 ×  {624} cm³
= 3.14 × 16 ×  208 cm³ = 10449·92 cm³

 Therefore, the quantity of milk in the container =
1000 litres = 10·45 litres Cost of milk at the rate of Rs. 15 per litre = Rs. {10·45 × 15} = Rs.156·75

Surface area of the metal sheet used to make the container
=
= (3·14) × {20 × 28 + 64} cm²
= (3·14) ×  624 cm² 
= 1959·36 cm²
Therefore, the cost of the metal sheet at rate of Rs. 5 per 100 cm²

=  
 = Rs. 97·97 approx.

Ex. 3. The height of a cone is 40 cm. A small cone is cut off at the top by a plane parallel to the base. If the volume of the small cone be 1/64 of the volume of the given cone, at what height above the base is the section made ?

Sol. Let R be the radius of the given cone, r the radius of the small cone, h be the height of the frustum and h₁ be the height of the small cone.
In figure 13.49, ΔONC and ΔOMA are similar (ΔONC ~ DOMA)

  
              ...(i)

We are given that

From (i) and (ii)

Therefore, h = 40 – h₁ = (40 – 10) cm ⇒ h = 30 cm Hence, the section is made at a height of 30 cm above the base of the cone.