A sphere is a solid generated by the revolution of a semicircle about its diameter as axis.

**The Sphere**

A sphere is a solid generated by the revolution of a semicircle
about its** **diameter as axis.

Every plane section of a sphere is a circle. The line of section
of a sphere by a plane passing through the centre of the sphere is called a
great circle; all other plane sections are called small circles.

As shown in the diagram, circle with *CD* as diameter is a
great circle, whereas, the circle with *QR* as diameter is a small circle.

Place a sphere inside a right circular cylinder of equal diameter
and height. Then the height of the cylinder will be the diameter of the sphere.
In this case, Archimedes proved that the outer area of the sphere is same as
curved surface area of the cylinder.

That is, Surface area of sphere =curved surface area of cylinder

= 2πrh = 2*πr *(2*r*)

Surface area of a sphere = 4*πr*^{2} sq.units

A section of the sphere cut by a plane through any of its great
circle is a hemisphere.

By doing this, we observe that a hemisphere is exactly half the
portion of the sphere.

Curved surface area of hemisphere =

C.S.A. of a hemisphere = 2*πr*^{2} sq.units

Total surface area of hemisphere = C.S.A. +Area of top circular
region

= 2*πr *^{2} +*πr*^{2}

T.S.A. of a hemisphere = 3*πr*^{2} sq.units

Let the inner radius be *r* and outer radius be *R*,

then thickness = *R−r*

Therefore, C.S.A. = Area of external hemisphere + Area of
internal hemisphere

= 2*πr*^{2} + 2*πr*^{2}

C.S.A. of a hollow hemisphere = 2*π*(*R*^{2}
+ *r*^{2} ) sq. units

T.S.A. = C.S.A. + Area of annulus region

= 2 π (*R* ^{2} + *r*^{2} ) + π (*R*
^{2} −*r*^{2} )

= π [2*R*^{2} + 2*r*^{2} + *R*^{2}
− *r*^{2}]_{}

T.S.A. of a hollow hemisphere = *π*(3*R*^{2} +
*r*^{2} ) sq. units

Find the diameter of a sphere whose surface area is 154 m^{2}.

** **Let

Given that, surface area of sphere = 154 m^{2}

4*πr*^{2 }= 154

Therefore, diameter is 7 m

**Example 7.9**

The radius of a spherical balloon increases from** **12** **cm to** **16** **cm as air** **being pumped into it.
Find the ratio of the surface area of the balloons in the two cases.

*Solution*

Let *r*_{1} and *r*_{2 }be the
radii of the balloons.

Therefore, ratio of C.S.A. of balloons is 9:16.

If the base area of a hemispherical solid is** **1386** **sq. metres, then find
its total surface area?

Let* **r*** **be the radius of the hemisphere.

Given that, base area = *πr*^{2} = 1386 sq. m

T.S.A. = 3*πr*^{2} sq.m

= 3 ×1386 = 4158

Therefore, T.S.A. of the hemispherical solid is 4158 m^{2 }

For finding the C.S.A. and T.S.A. of a hollow sphere, the formulla for finding the surface area of a sphere can be used.

**Example 7.11 **The internal and external radii of a hollow
hemispherical shell are** **3** **m and** **5 m respectively. Find the T.S.A. and C.S.A. of the shell.

** Solution **Let the internal and external radii of the
hemispherical shell be

Given that, *R* = 5 m, *r* =3 m

C.S.A. of the shell = 2*π*(*R*^{2} + *r*^{2}
) sq. units

= 2 × (22/7) × (25 + 9) = 213.71

T.S.A. of the shell = *π* (3*R*^{2} + *r*^{2}
) sq. units

= (22/7) × (25 + 9) = 264

Therefore, C.S.A. = 213.71 *m*^{2} and T.S.A. =
264 m^{2}.

A sphere, a cylinder and a cone (Fig.7.20) are of the same radius,
where as cone and cylinder are of same height. Find the ratio of their curved
surface areas.

Required Ratio** **=

=
4*π**r*^{2}
: 2*π**rh* : *π**rl* , (*l* = √[*r* ^{2} + *h*^{2}
] = √[2*r*^{2}]
= √2*r* units)

=
4 : 2: √2 = 2√2 : √2 : 1

Tags : Definition, Formula, Solved Example Problems | Mensuration | Mathematics , 10th Mathematics : Mensuration

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10th Mathematics : Mensuration : Surface Area: The Sphere | Definition, Formula, Solved Example Problems | Mensuration | Mathematics

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