NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.9 – Part 2

Curved surface area, total surface are and volumes of common solids

Curved Surface Area, Total Surface Are and Volumes

Curved surface area, total surface are and volumes of common solids

Q-2 The front compound wall of a house is decorated by wooden spheres of diameter 21cm, place on small supports as shown in the figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5cm and height 7cm and is to be painted black. Find the costs if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2 .

Front compound wall of house,give it diameter 21cm,radius=1.5cm,height=7cm

Front Compound Wall of House

Front compound wall of house,give it diameter 21cm,radius=1.5cm,height=7cm

Solution:

Given,

  • Compound wall of a house is decorated by wooden spheres of diameter 21cm .

  • Cylinder has radius 1.5cm

  • Height of cylinder 7cm

  • Cost of black paint (for spheres) is 5 paise per cm2

  • Cost of silver paint (for cylinder) is 2 5 paise per cm2

Surface area to be silver painted =8 (curved surface area of the sphere - area of circle on which sphere is resting)

  • 8(4πR2πr2)cm2

  • 8π(4×227×212×212227×1.52) ( R=212andr=1.5 )

  • 8π(4×44142.25)

  • 8π(4412.25)

  • 8π(438.75)cm2

Cost of silver paint =25 paise per cm2 , therefore total cost of silver painting,

  • =Rs.(8×227×438.75×25100)

  • =Rs.(193057)

  • =Rs.2757.86 (approx.)

Surface area to be black painted =8× curved area of cylinder. (since there are 8 cyliners)

  • 8×2πrh

  • 8×2×227×1.5×7cm2

  • 528cm2

Cost of black paint =5 paise per cm2 . Therefore total cost of black painting,

  • =Rs(528×5100)

  • =Rs26.4

Therefore, total cost of painting =Rs(2757.86+26.4) = Rs2784.26 (approx.)

Q-3 The diameter of a sphere is decreased by 25%, By what percent does its curved surface area decrease?

Solution:

Let d be the diameter of the circle,

Its surface area = 4πr2 , since r=d2 . Therefore, surface area = 4π(d2)2= 4πd24=πd2

On decreasing its diameter by 25% , the new diameter becomes, d'=(75100×d)=3d4

Therefore, surface area =4π(d'2)2=4π(12×3d4)2=4π(3d28)2=4π9d264=πd2916

Therefore decrease in surface area between two spheres = πd2πd2×916

  • =πd2(1916)

  • =πd2(716)

Therefore percentage decrease in surface area =(decreaseinareainitialsurfacearea×100)%

  • =(πd2×716πd2×100)%

  • =πd2×7πd2×16×100%

  • =(70016)%

  • =43.75%

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