# NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.9 – Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q-2 The front compound wall of a house is decorated by wooden spheres of diameter 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 and height and is to be painted black. Find the costs if silver paint costs paise per and black paint costs paise per .

Solution:

- Given,
- Compound wall of a house is decorated by wooden spheres of diameter .
- Cylinder has radius
- Height of cylinder
- Cost of black paint (for spheres) is paise per
- Cost of silver paint (for cylinder) is 2 paise per

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

- ()

Cost of silver paint paise per , therefore total cost of silver painting,

- (approx.)

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

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

Therefore, total cost of painting = (approx.)

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

Solution:

Let *d* be the diameter of the circle,

Its surface area = , since . Therefore, surface area =

On decreasing its diameter by , the new diameter becomes,

Therefore, surface area

Therefore decrease in surface area between two spheres =

Therefore percentage decrease in surface area