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NCERT Solutions for Class 9 Maths Exercise 13.9 Chapter 13 -Surface areas and Volumes(Term 2)
NCERT has exactly planned the Solutions in such a manner so that it shows you the different ideas of Geometry alongside the use of pertinent formulas. NCERT Solutions For Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.9 helps in tackling the application-level complex issue in a simple manner.
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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Term 2 )CBSE Board exam
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NCERT Solutions for Class 9 Maths Exercise 13.9 Chapter 13 -Surface areas and Volumes(Term 2)
Q1. A wooden bookshelf has external dimensions as follows: Height = 110cm, Depth = 25cm,
Breadth = 85cm (see fig. 13.31). The thickness of the plank is 5cm 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 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
Ans. The external dimensions of the wooden bookshelf are
Height=110 cm, breadth =25 cm and depth of the wooden bookshelf is 85 cm
Let l=85 cm,b =25 cm and h=110 cm
Since thickness of the plank is 5 cm everywhere
Internal length(L) of front portion =85 -2×5 =85 -10 =75 cm
There are 4 planks in front portion
Total internal height of three self =110 -4×5 =90 cm
Hight(H) of single self =90/3 =30 cm
Breadh(B) of each self = 25 -5 =20 cm
Total external surface area of the bookself to be polished
= TSA of whole of bookself(as a cuboid) – 3×area of open portion of the bookself
=2(lb +bh +hl) – 3(length of each self×height of each self)
=2(85×25 +25×110+110×85) – 3(75×30)
=2(2125 +2750 +9350) – 3×2250
=2×14225 – 3×6750
=28450 – 6750
=21700
External Area to be polished is 21700 cm²
Cost of polishing is 20 paise per cm²
20 paise =Rs 0.20
Therefore cost of polishing external surface area of the bookself
= 21700×0.20
=4340
So,the cost of polishing is Rs 4340
Area of the bookshelf to be painted
= 3×TSA of single shelf(as a cuboid) -3×open area of each self
=3×2(LB +BH+HL) – 3(L×H)
=6(75×20+20×30 +30×75) -3(75×30)20100-6750
=6(1500 +600 +2250) -3(2250)
=6×4350 -6750
=26100-6750
=19350
Area to be painted is 19350 cm²
The cost of painting is 10 painse per cm²
10 painse =Rs 0.10
Therefore cost of painting is 0.10 ×19350 =Rs 1935
Total expenses required for polishing and painting the surface of the bookshelf
= Cost of polishing + Cost of painting
=Rs 4340 +Rs 1935
=Rs 6275
Hence total expenses of the painting and polishing is Rs 6275
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Q2.The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in fig. 13.32. 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 cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
Ans. The diameter of the wooden sphere is 21 cm
The radius(r) of the wooden sphere is 21/2 =10.5 cm
Surface area of the sphere
=4πr²
=4(22/7)×10.5×10.5
= 88 ×1.5×10.5
=1386
Surface area of a single sphere is 1386 cm²
Area of the sphere to be painted silver =Surface area of the sphere -area of lower circular end of the sphere attached to cynder
= 1386 -πr²= 1386 -(22/7)×1.5×1.5 =1386 -7.07=1378.93
The surface area of 8 spheres to be painted =8×1378.93=11031.44 cm²
The cost of silver paint is 25 paise per cm²
35 paise =Rs 0.25
The cost of painting silver colour is
=0.25× 11031.44=Rs 2757.86
Area of the cylinder to be painted
= Curved surface area of the cylinder
=2πrh, where radius (r) of the cylinder is given 1.5 cm and height(h) is given 7 cm
=2(22/7)×1.5×7
= 44×1.5
=66
The surface area of 8 cylinders to be painted is 8×66 =528 cm²
The cost of black paint is 5 paise per cm²
5 paise =Rs 0.05
The cost of painting black colour is
=0.05×528
=Rs 26.40
The total cost of painting both colours
=Rs 2757.86 +Rs 26.40
= Rs 2784.26
Q3.The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Ans. Let diameter of the sphere is 2r then its radius is r
Its surface area is,S =4πr²
Since diameter of the sphere is decreased by 25% then diameter of modified sphere is
=2r -25% of 2r
= 2r – r/2
= (4r -r)/2 =3r/2
So, radius of the modified sphere is 3r/4
Surface area of the modified sphere is
S’=4π(3r/4)²=9πr²/4 cm²
The curved surface area of the sphere is decreased by
S -S’ =4πr² – 9πr²/4 =7πr²/4
The percent by which the curved surface area of the sphere is decreased
=[(S-S’)/S]×100
=6.25 ×7
=43.75 %
Hence curved surface area of the sphere is decreased by 43.75 %
The volume of the hemispherical bowl is 22.46 mm³
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