**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 Mathsย 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.ย 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 cm ^{2}ย and the rate of painting is 10 paise per cm^{2}, 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 -2ร5 =85 -10 =75 cm

There are 4 planks in front portion

Total internal height of three self =110 -4ร5 =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) – 3รarea of open portion of the bookself

=2(lb +bh +hl) – 3(length of each selfรheight of each self)

=2(85ร25 +25ร110+110ร85) – 3(75ร30)

=2(2125 +2750 +9350) – 3ร2250

=2ร14225 – 3ร6750

=28450 – 6750

=21700

**External Area to be polished is 21700 cmยฒ**

Cost of polishing is 20 paise per cmยฒ

20 paise =Rs 0.20

Therefore cost of polishing external surface area of the bookself

= **21700ร0.20**

=4340

**So,the cost of polishing is Rs 4340**

Area of the bookshelf to be painted

= 3รTSA of single shelf(as a cuboid) -3รopen area of each self

=3ร2(LB +BH+HL) – 3(LรH)

=6(75ร20+20ร30 +30ร75) -3(75ร30)20100-6750

=6(1500 +600 +2250) -3(2250)

=6ร4350 -6750

=26100-6750

=19350

**Area to be painted is 19350 cmยฒ**

The cost of painting is 10 painse per cmยฒ

10 painse =Rs 0.10

**Therefore cost of painting is 0.10 ร19350 =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 cm ^{2}ย and black paint costs 5 paise per cm^{2}.**

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

=4ฯrยฒ

=4(22/7)ร10.5ร10.5

= 88 ร1.5ร10.5

=1386

Surface area of a single sphere is 1386 cmยฒ

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 -ฯrยฒ= 1386 -(22/7)ร1.5ร1.5 =1386 -7.07=1378.93

**The surface area ofย 8 spheres to be painted =8ร1378.93=11031.44 cmยฒ**

The cost of silver paint is 25 paise per cmยฒ

35 paise =Rs 0.25

The cost of painting silver colour is

=0.25ร 11031.44=Rs 2757.86

Area of the cylinder to be painted

= Curved surface area of the cylinder

=2ฯrh, where radius (r) of the cylinder is given 1.5 cm and height(h) is given 7 cm

=2(22/7)ร1.5ร7

= 44ร1.5

=66

**The surface area ofย 8 cylinders to be painted is 8ร66 =528 cmยฒ**

The cost of black paint is 5 paise per cmยฒ

5 paise =Rs 0.05

The cost of painting black colour is

=0.05ร528

=Rs 26.40

**The total cost of painting both coloursย **

=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 =4ฯrยฒ

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’=4ฯ(3r/4)ยฒ=9ฯrยฒ/4 cmยฒ

The curved surface area of the sphere is decreased by

S -S’ =4ฯrยฒ – 9ฯrยฒ/4ย =7ฯrยฒ/4

The percent by which the curved surface area of the sphere is decreased

=[(S-S’)/S]ร100

=6.25 ร7

=43.75 %

**Hence curved surface area of the sphere is decreased by 43.75 %**

**The volume of the hemispherical bowl is 22.46 mmยณ**

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Chapter 1-Sets | Chapter 9-Sequences and Series |

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Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

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