NCERT Solutions for Surface Areas and Volumes Exercise 13.9 Class 9 Maths

Exercise 13.9 Class 9 Maths NCERT Solutions: In this article, we have covered the topic Volume of a Sphere, which is Exercise 13.9 of Class 9 Mathematics. The NCERT Solutions for Class 9 Maths Chapter 13 deals with Surface Areas and Volumes exercises. The subject-matter experts at Embibe have solved the NCERT solutions by keeping in mind the understanding and grasping abilities of the students. All the solutions for Class 9 Maths subjects are according to the NCERT syllabus and CBSE guidelines.

Each question of Ex 13.9 Class 9 is solved in a straightforward way to help the students understand and prepare well for the final exams. Surface Areas and Volumes is an important topic and will help students score good marks in the final exam. This forms the basics for Class 10th and students must know the formulas and their applications in problems thoroughly. Read on to access the Class 9 Maths Exercise 13.9 Solutions PDF and other related information.

NCERT Solutions for Surface Areas and Volumes Exercise 13.9 Class 9 Maths: Overview

We have provided the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 to help students to learn and perform well in the exam. These answers will help them clear all doubts about the volume of a sphere and hemisphere. Students will help to get the basics of the Surface Areas and Volumes to understand its advanced concepts.

However, Embibe suggests that students should first try to solve the textbook exercise questions independently and if they find any issues, they can refer to these solutions. Let us have an overview of the Class 13 Maths chapter before going into the details of Class 9 Maths Exercise 13.9 Solutions:

NCERT Solutions for Class 9 Maths Chapter 13 Ex 13.9: Download PDFs

The Volume of a Sphere solution is available in PDF format for students so that they can download it and study offline. Many students may not have access to a trustworthy internet connection, so they can download the PDF and keep a hard copy for reference. We suggest students to bookmark this page for future reference.

Chapter 13 of Class 9 covers the various concepts of Surface Areas and Volumes and formulas to solve all the questions. Students must understand this chapter completely to score better marks in future classes. By referring to this PDF students can understand the steps to solve the questions by downloading this PDF solution. Check the images below to find the solution to the Class 9 Maths Exercise 13.9.

Important Questions for NCERT Solutions Class 9 Maths Chapter 13 Exercise 13.9

Exercise 13.9 is optional but still solving them will boost the confidence of students to appear in the final exam. There are only 3 questions and we have given the text format for the reference. Candidates can check the problems and solutions here:

Q.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 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. [ write only numerical value without rupee symbol]

6275

Solution: The outer dimension of the bookshelf is

Length (l) =85 cm; Breadth (b) =25 cm; Height (h) =110 cm; Thickness of the plank =5 cm; External surface area of shelf while leaving out the front face of the shelf =lh+2(lb+bh) =85×110+285×25+25×110 =19100 cm²

The inner dimension of the shelf is
Height (h1)=110-10=100 cm

Width b1 =25-10=15 cm Length l1 =85-10=75 cm
Area of the front face =(l×h)-(l1×h1)+2l1×width =85×110-75×100+275×5 =2600 cm². Area to be
polished =19100+2600=21700 cm².

Length, breadth and height of each row are l2=85-10=75 cm, b2=25-5=20 cm and h2=110-10-5-5÷3=30 cm respectively.

Area to be painted in 1 row
=2(l2+h2)b2+l2h2 =275+30×20+75×30 cm2 =4200+2250 cm2 =6450 cm². Area to be painted in 3 rows =3×6450=19350 cm².
Cost of polishing the area of 1 cm²=₹ 0.20

Cost of polishing the area of 21700 cm²=21700×0.20=₹4340
Cost of painting area of1 cm²=₹ 0.10 Cost of painting the surface of 19350 cm²=₹19350×0.1=₹1935. Total expenses needed for painting and polishing =4340+1935=₹6275. Hence, the total expenses required for polishing and painting the surface of the bookshelf is ₹6275.

Q.2: The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in 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. If the cost of paint required is ₹k (correct up to two decimal places), if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm² then find k. [Take π=227].

2784.26

Solution: Given,

Diameter of the wooden block =21 cm
Radius of the wooden sphere (r)=212 cm=10.5 cm We know that, Surface area of the wooden sphere =4πr² =4×227×10.52 cm²=1386 cm² Therefore, the
surface area of the sphere =1386 cm²
Also, given that, the radius of the cylindrical support =1.5 cm
Height of cylindrical support =7 cm
So, the curved surface area of a cylindrical support =2πrh =2×227×1.5×7 cm²=66 cm²
Area of the circular end =πr²
=227×1.52 cm²
=7.07 cm2 (approx.) Area to be painted silver =8×1386-7.07 cm² =8×1378.93 cm²=11031.44 cm²
Cost of painting with the silver colour =₹ 11031.44×0.25=₹ 2757.86.
Area to be painted black =8×66 cm²=528 cm²
The cost of painting with black colour =528×0.05=₹ 26.40 Hence, the total cost is ₹2757.86+₹26.40=₹2784.26. Therefore, the cost of paint required is ₹2784.26.

Q.3: The diameter of a sphere is decreased by 25%. If its curved surface area decreases by k% then find k in decimal form correct up to two places.

43.75

Solution: Let originally the diameter of the sphere be 2r.
Then, radius of the sphere = r
Surface area of the sphere = 4πr2. New diameter of the sphere = 2r – 2r ×25100=3r2. New radius of the sphere =3r4
Surface area of the new sphere = 4π3r42=9πr24
Decrease in surface area = 4πr2-9πr24=7πr24
Thus, the percentage decrease
=7πr24×1004πr2=1754=43.75%
Hence, the curved surface area of the sphere decreases by 43.75%. Thus, k=43.75

Students may find the direct PDF links for Exercise 13.9 Class 9 Maths below.

DOWNLOAD NCERT MATHS CLASS 9 CHAPTER 13 EXERCISE 13.9

Benefits of Studying from Embibe’s NCERT Solutions for Class 9 Maths Chapter 13

Check the list of benefits that students can gain after solving NCERT solutions at Embibe:

• Embibe’s Mathematics SMEs create NCERT Maths Class 9 Solutions, and they provide a deep understanding of the Volume of a sphere.
• Each question of Ex 13.9 Class 9 is solved in a simple manner with a step-by-step guide for students to get a good hold of the concepts and formulas.
• Class 9 Maths Chapter 13 Exercise 13.9 is solved in a more straightforward way to help students understand the applications of formulas. This will, in turn, help them solve problems with ease in the exam.
• A full assessment of topics and important questions are provided, along with marks and weightage. These features will assist students in Class 9 exam preparation.
• Students can use these NCERT solutions for last-minute revisions and they can get rid of doubts and confusion.
• Students can also prepare for their future entrance exams as these concepts form the basics.
• The NCERT Class 9 Maths Solutions are provided in the form of PDFs for students to access for and use while studying offline.
• Along with deep analysis and understanding of the Class 9 Maths Exercise 13.9 Solutions PDF, students are also provided with the chapter-wise mock tests to improve their exam-taking skills.
• Embibe will provide feedback analysis at the end of each chapter-wise mock test. Through this, students can understand their mistakes and rectify them to score good marks.

About NCERT Solutions for Class 9 Maths Ex 13.9

NCERT Class 9 Maths Chapter 13 Exercise 13.9 discusses the Volume of a sphere. In this exercise, students can learn about the formulas related to the Volume of a sphere, hemisphere and their applications in various problems. Let us see some important points discussed in Exercise 13.9 Class 9 Maths:

• The volume of a sphere and its application
• The volume of a hemisphere and its usage

NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 Chapter Description

In this exercise, we have discussed the problems related to the volume of a sphere and hemisphere. The concepts covered in Class 9 Exercise 13.9 include the Volume of a sphere, hemisphere and its formula. By solving the NCERT solutions, students will also know which value of π is used to solve a particular problem.

Students will learn the following formulas in Class 9 Maths Chapter 13:

• Surface area of a cuboid = 2(lb + bh + hl)
• Surface area of a cube = 6a²
• Curved surface area of a cylinder = 2πrh
• Total surface area of a cylinder = 2πr(r + h)
• Curved surface area of a cone = πrl
• Total surface area of a right circular cone = πrl + πr² , i.e., πr (l + r)
• Surface area of a sphere of radius r = 4 πr²
• Curved surface area of a hemisphere = 2πr²
• Total surface area of a hemisphere = 3πr²
• Volume of a cuboid = l × b × h
• Volume of a cube = a³
• Volume of a cylinder = πr²h
• Volume of a cone = 1/3 πr²h
• Volume of a hemisphere = 2/3 πr³
• Volume of a sphere = 4/3 πr³
  [Here, letters l, b, h, a, r, etc., have been used in their usual meaning, depending on the context.]

FAQs on NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9

Here are some of the frequently asked questions related to Class 9 Maths Chapter 13 Exercise 13.9 NCERT Solutions and their answers:

Related Links:

We have provided you with detailed NCERT Solutions for Class 9 Maths Chapter 13. So, you can quickly get the NCERT Maths Class 9 Solutions PDF for. Download the PDF and start your preparation. Do not forget that the best way to prepare for Maths is to practice all the textbook questions. Students can also download Class 9 NCERT Solutions, Books and syllabus for all the subjects from the Embibe.

Stay tuned to embibe.com for more on NCERT solutions.