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AP Board Class 9 Maths Chapter 10 Surface Areas and Volumes Ex 10.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 10 Surface Areas and Volumes Ex 10.4 Book Answers

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Andhra Pradesh Board Class 9th Maths Chapter 10 Surface Areas and Volumes Ex 10.4 Textbooks Solutions PDF

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Andhra Pradesh State Board Class 9th Maths Chapter 10 Surface Areas and Volumes Ex 10.4 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths
Chapters	Maths Chapter 10 Surface Areas and Volumes Ex 10.4
Provider	Hsslive

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AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.4

Question 1.
The radius of a sphere is 3.5 cm. Find its surface area and volume.
Solution:
Radius of the sphere, r = 3.5 cm

Question 2.
The surface area of a sphere is 101827 cm² . What is its volume ?
Solution:
Surface area of sphere = 4πr²
= 101827 cm²

= 3054.857cm³
≅ 3054.86cm³

Question 3.
The length of equator of the globe is 44 cm. Find its surface area.
Solution:
Length of the equator of the globe 2πr = 44 cm.
2 × 227 × r = 44
∴ r = 44×72×22 = 7cm
∴ surface area = 4πr²
= 4 × 227 × 7 × 7
= 4 × 22 × 7
= 616cm²

Question 4.
The diameter of a spherical ball is 21 cm. How much leather is required to prepare 5 such balls?
Solution:
Diameter of the spherical ball d’ = 21 cm
Thus, its radius r = 𝑑2=212 = 10.5 cm
Surface area of one ball = 4πr²
= 4 × 227 × 10.5 × 10.5
= 88 × 1.5 × 10.5 = 1386 cm²
∴ Leather required for 5 such balls
= 5 × 1386 = 6930 cm²

Question 5.
The ratio of radii of two spheres is 2 : 3. Find the ratio of their surface areas and volumes.
Solution:
Ratio of radii r₁ : r₂ = 2 : 3
Ratio of surface area
= 4πr₁² : 4πr₂²
= 2²: 3² = 4 : 9
Ratio of volumes
= 4/3 πr₁³ : 4/3 πr₂³
= 2³ : 3³ = 8 : 27

Question 6.
Find the total surface area of hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
Radius of the hemisphere = 10 cm
Total surface area of the hemisphere = 3πr²
= 3 × 3.14 × 10 × 10
= 9.42 × 100
= 942 cm²

Question 7.
The diameter of a spherical balloon increases from 14 cm. to 28 cm. as air is being pumped into it. Find the ratio of surface areas of the balloons in the
two cases.
Solution:
The diameter of the balloon, d = 14 cm
Thus, its radius, r = 𝑑2=142 = 7 cm
∴ Surface area = 4πr² = 4 × 227 × 7 × 7
= 88 × 7 = 616cm²
When air is pumped, the diameter = 28 cm
thus its radius = 𝑑2=282 = 14 cm
Its surface area = 4πr²
= 4 × 227 × 14 × 14
= 88 × 28 = 2464 cm²
Ratio of areas = 616 : 2464
= 1 : 4

(OR)

Original radius = 142 = 7 cm
Increased radius = 282 = 14cm
Ratio of areas = r₁² : r₂²
= 7² : 14²
= 7 × 7 : 14 × 14
= 1:4

Question 8.
A hemispherical bowl is made of brass, 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the ratio of outer surface area to inner surface area.
Solution:
Inner radius of the hemisphere ‘r’ = 5 cm
Outer radius of the hemisphere ‘R’
= inner radius + thickness
= (5 + 0.25) cm = 5.25 cm
Ratio of areas = 3πR²: 3πr²
= R² : r²
= (5.25)²: 5²
= 27.5625 : 25
= 1.1025:1
= 11025 : 10000
= 441 : 400
[Note : If we read “radius as diameter” then we get the T.B. answer]

Question 9.
The diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34 g/c³. What is the, weight of the ball ?
Solution:
The diameter of the ball = 2.1 cm
Thus, its radius, r = 𝑑2=2.12 = 1.05 cm
Volume of the ball V’ = 43πr³
= 43×227 x 1.05³ = 101.8721
∴Weight of the ball = Volume × density
= 4.851 × 1.34
= 55.010

Question 10.
A metallic cylinder of diameter 5 cm 1 and height 3 13 cm is melted and cast into a sphere. What is its diameter ?
Solution:
Diameter of the cylinder’d’ = 5 cm
Thus, its radius, r = 𝑑2=52 = 2.5 cm
Height of the cylinder,

Volume of the cylinder

Given that cylinder melted to form sphere
∴ Volume of the sphere = Volume of the cylinder

(Where r is the radius of the sphere)
r³ = 34 × 2.5 × 2.5 × 103
r³ = 2.5³
∴ r = 2.5 cm
Hence its diameter, d = 2r
= 2 × 2.5 = 5 cm

Question 11.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold ?
Solution:
Diameter of the hemispherical bowl = 10.5 cm
Thus its radius = 𝑑2=10.52 = 5.25cm
Quantity of milk, the bowl can hold = Volume of the bowl = 23πr³
= 23×227 × 5.25 × 5.25 × 5.25
= 303.1875 cm³
= 303.18751000 lit = 0.303 lit.

Question 12.
A hemispherical bowl has diameter 9 cm. The liquid is poured into cylindrical bottles of diameter 3 cm and height 3 cm. If a full bowl of liquid is Riled in the bottles, find how many
bottles are required ?
Solution:
Diameter of the hemispherical bowl ‘d’ = 9 cm
Its radius, r = 𝑑2=92 = 4.5cm
Volume of its liquid = Volume of the bowl = 23 πr³
= 23×227 × 4.5 × 45 × 4.5
Diameter of the cylindrical bottle, d = 3 cm
Its radius, r = 𝑑2
= 3.02
= 1.5cm

Height of the bottle, h = 3 cm
Let the number of bottles required = n
Then total volumes of these n bottles = n πr²h
But this is equal to volume of the bowl
Hence n. 227 × 1.5 × 1.5 × 3
= 23×227 × 4.5 × 4.5 × 4.5
∴ n = 23×20.251.5 = 9
∴ Number of bottles required = 9

AP Board Textbook Solutions PDF for Class 9th Telugu

Andhra Pradesh Board Class 9th Maths Chapter 10 Surface Areas and Volumes Ex 10.4 Textbooks for Exam Preparations

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