Surface Areas and Volumes – Exercise 13.9 – (MATHEMATICS) – 9th Class

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Exercise 13.9 

1.  A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 poise per cm², find the total expenses required for polishing and painting the surface of the bookshelf.

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Sol. Area of external faces to be polished
= Total surface area of the cuboid
— 3 × area of one rectangular (open) surface

= [2 (85 × 110 + 110 × 25 + 85 × 25) — 3 × 30 × 75] cm²

= [2(9350 + 2750 + 2125) – 6750] cm² = 21700 cm².

Dimensions of inner boxes are 75 cm × 30 cm × 20 cm
∴ Area of 3 inner boxes to be painted
= 3 × [2(75 + 30) × 20 + 75 × 30] cm²

= 3 [4200 = 2250] cm2

= 19350  cm2

Total expenses for polishing and painting

= Rs.\left[ {{{20} \over {100}} \times 21700 + {{10} \over {100}} \times 19350} \right]

= Rs.(4340 + 1935) = Rs.6,275.


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

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Sol.  Surface area of 8 spheres

 = 8 \times \left[ {4 \times {{22} \over 7} \times {{\left( {{{21} \over 2}} \right)}^2} - {{22} \over 7} \times {{(1.5)}^2}} \right]

= 11031.43 cm².

Curved surface area of 8 support cylinders

 = 8 \times 2 \times {{22} \over 7} \times 1.5 \times 7c{m^2} = 528c{m^2}

∴   Cost of paint = Rs.\left[ {{{25} \over {100}} \times 11031.43 + {5 \over {100}} \times 528} \right]

= (2757.85 + 26.40) = Rs.2784.25.


3. The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

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Sol. Let diameter of the sphere = d units

⇒  Radius = {d \over 4} units.

∴  Surface area = 4\pi {\left( {{d \over 2}} \right)^2}sq.

units = πd2 sq. units.                                …..(i)

Diameter of new sphere = \left( {d - {{25} \over {100}}d} \right)units = {3 \over 4}d{\rm{ }}units.

Radius of new sphere = {3 \over 8}d units.

∴   Surface area of new sphere = 4\pi {\left( {{3 \over 8}d} \right)^2}

 = {9 \over {16}}\pi {d^2} sq. units.                   …….(ii)

Decrease = \left( {\pi {d^2} - {9 \over {16}}r{d^2}} \right)sq. units = {7 \over {16}}\pi {d^2} sq. units.

∴ Percentage decrease = {{{7 \over {16}}\pi {d^2}} \over {\pi {d^2}}} \times 100 = {{700} \over {16}}\%

=43.75%


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