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 

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 3 :

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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