Class 9 Maths


Area Volume

Exercise 13.9

Question 1: A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm. 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 sq cm and the rate of painting is 10 paise per sq cm, find the total expenses required for polishing and painting the surface of the bookshelf.

cuboid

Answer: Area of rear wall = 110 × 85 = 9350 sq cm

Area of side walls = 2 × 110 × 25 = 5500 sq cm

Area of base and roof = 2 × 85 × 25 = 4250 sq cm

Area to be polished on front can be calculated as follows:

Area of vertical (blue) portion = 2 × 110 × 5 = 1100 sq cm

Area of horizontal (blue) portion = 4 × (85 – 10) × 5 = 1500 sq cm

Total Area to be polished = 9350 + 5500 + 4250 + 1500 + 1100 = 21700 sq cm

Cost = 21700 × 20 = 434000 paise = Rs. 4340

Area of any one of inner shelves = 2(30 + 75) × 25

(Note: If you deduct the thickness of 4 planks (4 × 5 = 20 cm) from height (110 cm), you will get 90 cm as remainder. Dividing this figure by 3 gives the height of each shelf as 30 cm. Similarly, subtracting 10 cm from width (85 cm) gives 75 cm as remainder.)

= 2 × 105 × 25 = 5250 sq cm

Area of three shelves = 3 × 5250 = 15750 sq cm

Cost = 15750 × 10 = 157500 paise = Rs. 1575

Total cost = 4340 + 1575 = Rs. 5915

Question 2: The front compound wall of a house is decorated by wooden sphere of diameter 21 cm, placed on small supports as shown in this 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 sq cm and black paint costs 5 paise per sq cm.

sphere

Answer: Surface Area of Sphere `=4πr^2`

`=4xx(22)/7xx(21)/2xx(21)/2=1386` sq cm

We need to deduct the area of base of cylinder from area of sphere because paint can be applied only on remaining area of sphere.

Area of base of cylinder `=pi;r^2`

`=(22)/7xx1.5^2=6.6` sq cm

Area to be painted on sphere = 1386 – 6.6 = 1379.4 sq cm

Cost of silver painting 8 spheres `=8xx1379.4xx0.25` = Rs. 2658.80

Curved surface area of cylinder `=2πr\h`

`=2xx(22)/7xx1.5xx7=66` sq cm

Cost of black paint on 8 cylinders `=8xx66xx0.05` = Rs. 26.40

Total cost = 2658.80 + 26.40 = Rs. 2685.20

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

Answer: 25% reduction in diameter means 25% reduction in radius as well. If radius of sphere is r then reduced radius is 0.75 r.

Surface area of sphere `=4πr^2`

Surface area of smaller sphere `=4π(0.75r)^2`

difference `=4πr^2-4π(0.75r)^2`

`=4πr^2-4π\xx0.5625r^2`

`=4πr^2(1-0.5625)`

`=4π\xx0.4375r^2`

This means, reduction in surface area = 43.75%