CURVED SURFACE AREA AND TOTAL SURFACE AREA OF SPHERE AND HEMISPHERE

Curved Surface Area and Total Surface Area of Hemisphere

Curved surface area 

=  2Πr2

Hollow hemisphere

= 2Π(R2 + r2) sq. units

Total surface area of 

=  3Πr2

Hollow hemisphere 

=  Π(3R2 + r2) sq. units

Problem 1 :

The radius of a sphere increases by 25%. Find the percentage increase in its surface area.

Solution :

Radius of hemisphere  =  r

Increased percentage of radius  =  125% of r

  =  125r/100

  =  5r/4

Surface area of hemisphere   =  2Πr2

Surface area of increased hemisphere   =  2Π(5r/4)2

  =  25Πr2/8

  =  [(25Πr2/8) - (2Πr2)]/(2Πr2)

  =  9Πr2/8(2Πr2)

  =  9Πr2/16 Πr2

  =  0.5625

  =  0.5625 (100)

  =  56.25%

Problem 2 :

The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost to paint the vessel all over at 0.14 per cm2 

Solution :

internal radius (r)  =  20/2  =  10 cm

External radius (R) =  28/2  =  14 cm

Curved surface area of hemisphere  = 2Π(R2 + r2

  =  2 (22/7)(142 + 102)

  =  2 (22/7)(196 + 100)

  =  1860.57

Cost of painting  =  0.14 per cm2 

Required cost  =  0.14 (1860.57)

  =  260.48 

Problem 3 :

What will be the diameter of a sphere whose surface area is 55.44 m2?

(a) 4.6 m (b) 4.2 m (c) 3.4 m (d) 2.4 m

Solution :

Surface area of sphere = 4Πr2

4Πr55.44

4 x (22/7) x r55.44

r= 55.44 x (1/4) x (7/22)

r= 4.41

r = √4.41

r = 2.1

Diameter = 2r

= 2 (2.1)

= 4.2 m

So, the required diameter is 4.2 m.

Problem 4 :

The curved surface area of a sphere of radius 7 cm is

Solution :

Curved surface area of the sphere = 4Πr2

Here radius (r) = 7 cm

= 4 x (22/7) x 7 x 7

= 616 cm2

Problem 5 :

Find the total surface area of a hemisphere of radius r/2 unit

Solution :

Total surface area of hemisphere = 3Πr2

radius = r/2

= 3 x Π x (r/2)2

= 3 x Π x (r2/4)

= (3/4) Π r

Problem 6 :

The radius of a spherical balloon increases from 7 cm to 14 cm when air is being pumped into it. Then find the ratio of surface area of the balloon in the two cases.

Solution :

Old radius = 7 cm and new radius = 14 cm

Surface area of balloon = 4Πr2

Let old radius be r and new radius be R.

= 4ΠR2 4Πr2

= 4Π(14)2 4Π(7)2

= (14)2 (7)2

= 4 : 1

So, the required ratio is 4 : 1.

Problem 7 :

A cylinder, a cone and a sphere are of the same radius and same height. Find the ratio of their curved surface.

Solution :

Radius of cylinder = radius of cone = radius of sphere = r = h

Surface area of cylinder = 2Πrh

Surface area of sphere = 4Πr2

Surface area of cone = Πrl

l = √r2 + h2

l = √r2 + r2

l = √2r2

l = r√2

Surface area of cylinder : Surface area of cone : Surface area of sphere

2Πrh : Πrl : 4Πr2

2r(r) : r (r√2) : 4r2

2r2 : r2√2 : 4r2

= 2 : √2 : 4

= 1 : √2 : 2

= 1 : 2 : 4

So, the required ratio is 1 : 2 : 4.

Problem 8 :

Find the radius of a sphere with a surface area of 1024π square inches.

Solution :

Surface area of sphere = 4Πr2

1024π = 4Πr2

1024 = 4r2

r2 = 1024/4

r2 = 256

r = 16

So, the radius of the sphere is 16 inches.

Problem 9 :

You friend claims that if the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is your friend correct? Explain your reasoning.

Solution :

Let radius of the sphere be r, then when it is double the new radius will be 2r.

Surface area of sphere = 4Πr2

When r = 2r

Surface area of new sphere = 4Π(2r)2

= 16Πr2

= 4 (4Πr2)

= 4(surface area of old sphere)

So, it is incorrect.

Curved Surface Area and Total Surface Area of Frustum Cone

C.S.A. of a frustum

  =  Π(R +r)l sq. units

T.S.A. of a frustum

  =  Π(R + r)l + Π(R2+ r2)

where, l = √[h2 +(R −r)2]

Problem 10 :

Th e frustum shaped outer portion of the table lamp has to be painted including the top part. Find the total cost of painting the lamp if the cost of painting 1 sq.cm is 2.

Solution :

Curved surface area of lamp  =  Π(R +r)l sq. units

R = 12 m, r = 6 m and h = 8 m

l = √[h2 +(R −r)2]

l = √[82 +(12 −6)2]

l = √[64 + 62]

l = √(64 + 36)  √100

l = 10

C.S.A of frustum shaped lamp  =  Π(R +r)l 

Area to be painted  =  Π(R +r)l + Πr2

  =  (22/7)[(12 + 6)(10) + 62]

=  (22/7)[180+36]

=  678.85

Cost per sq.cm is 2

Required cost  =  678.85 (2)

  =  1357.71

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