NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.9) Exercise 13.9
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Class 10 is a significant academic year for students. Their academic and professional endeavours are impacted by it. Success for a student, however, is not only determined by their performance over the course of a single academic year. The ninth grade is also considered a pivotal point in students’ academic careers since classes 9 and 10are two sides of the same coin. A student must pass Class 9 in order to qualify to take the Class 10 board exams. Additionally, Class 9 offers the outline of the structure of the curriculumfor subsequent classes. The likelihood of performing well on the Class 10 board exams is increased for students who are competent in Class 9 courses. Therefore, students must be persistentin their efforts with regard to Class 9.
There is a considerable likelihood that concepts covered in Class 9 will come up in competitive tests. Therefore, it is important for students to review the academic themes imparted to them in Class 9. The study of Science and Mathematics requires a lot of practice. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 are among the best resources for learning the subject matter. Students can use this resource to understand how well they understand the main ideas in Class 9 Mathematics and to improve their skills.
The Class 9 syllabus is important since it specifies the themes covered and sets the stage for the Class 10 board exams. As described in the syllabus, research requires a number of prerequisites and tasks. The Class 9 curriculum is critical for students who want to do well in their upcoming board exams. Mathematics experts at Extramarks advise students to learn Class 9 subjects sincerely since they may have an impact on their future academic careers. Whether the impact is positive or unfavourable will depend on students’ preparation for Class 9. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 and the Class 9 syllabus are available on the Extramarks website. To avail assistance for the Class 9 Mathematics exams, students can refer to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 . However, it is advised that students read the syllabus before beginning any subjectspecific preparation. Students can benefit from this practise by scheduling their preparation efficiently and saving a lot of time. When it comes to studying for exams, time management is essential.Students can improve their time management skills by using the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 along with other tools and thoroughly studying the curriculum.
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The National Council for Educational Research and Training publishes and distributes textbooks for school students across the country. The most remarkable thing about NCERT books for CBSE board exam preparation is that they focus on the conceptual foundation to help students understand the fundamental concepts. The NCERT texts are thorough and complete on their own, and the CBSE rarely asks students to study anything more outside these books.
The mentors of Extramarks advise students to use the NCERTrelated materials that are available on the Extramarks website, such as the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9, for a better understanding of NCERT textbooks. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 and other related resources can be used effectively for both test preparation and classroom studies.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.9) Exercise 13.9
Extramarks’ NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 are created by qualified mathematics mentors.On the Extramarks website, students can also access resources for primary and secondary grade levels that are prepared in a similar fashion to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9It is strongly recommended that students take advantage of resources such as the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 so that they fully comprehend everything covered in their classrooms. Students are unable to understand the middle and senior secondary school curriculum without a strong foundation in the learning materials used in elementary school classrooms. Students could become disinterested in the subject if there isn’t a strong foundation.. The use of tools developed by Extramarks’ expertise is encouraged to avoid reaching this stage. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9, among other resources, are easily accessible on Extramarks.
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Access NCERT Solutions for Class 9 Mathematics Chapter 13 – Surface Areas and Volumes
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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.9
The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 are the solutions to only one of the exercises from Class 9 Chapter 13.. The solutions are for Class 9 Maths Exercise 13.9.More of these resources are created by mentors who have been trained and educated in specific subjects.NCERT Solutions for Class 9
The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 are a reliable source of information for the students of Class 9 because these solutions provide them with stepbystep solutions to all the questions in Exercise 13.9. Students who want to practise the questions from this exercise should consult the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9.The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9 can be enough for them to practise as well as revise.CBSE Study Materials for Class 9
Mathematics is a subject that is part of everyday life, so it is important for students to understand it.. It can be a challenging subject for many students, but the major reason for that could be a weak conceptual foundation in the subject. Elearning resources available on the Extramarks website can help students form a strong foundation of Mathematics. Among these resources are the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9..
CBSE Study Materials
Students who wish to master the subject of Mathematics are advised to practise the mathematical questions as much as possible. They can do so with the help of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9. It is also important to prepare the theorems, formulas, and other theoretical components like definitions, etc. Without this, students might not be able to achieve a high score on the examination.
Q.1 A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see Figure). 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 cm^{2} 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
$\begin{array}{l}\text{Outer surface area}=2(110+85)\times 25+85\times 110\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\times 195\times 25+9350\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=19100\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Area vertical edges}=110\times 5+110\times 5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1100\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Area horizontal edges}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=75\times 5+75\times 5+75\times 5+75\times 5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=75(5+5+5+5)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=75\times 20\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1500\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Total external area}=19100\text{\hspace{0.17em}}{\mathrm{cm}}^{2}+1100\text{\hspace{0.17em}}{\mathrm{cm}}^{2}+1500\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=21700\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Internal length of each rack of bookshelf}\\ =8555\\ =75\text{\hspace{0.17em}}\mathrm{cm}\\ \text{Internal breadth of each rack of bookshelf}\\ =\frac{11020}{3}\\ =30\text{\hspace{0.17em}}\mathrm{cm}\\ \text{Internal depth of each rack of bookshelf}\\ =255\\ =20\text{\hspace{0.17em}}\mathrm{cm}\\ \text{Surface area of a bookshelf}\\ =2(75+30)\times 20+75\times 30\\ =2\times 105\times 20+2250\end{array}$ $\begin{array}{l}=4200+2250\\ =6450\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \mathrm{Surface}\text{area of three bookshelves}\\ =3\times 6450\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ =19350\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Rate of polishing external surface of bookshelf}\\ =\u20b9\text{\hspace{0.17em}\hspace{0.17em}}\frac{20}{100}\\ =\u20b9\text{\hspace{0.17em}}0.20{\text{per m}}^{\text{2}}\\ \mathrm{Rate}\text{of painting internal surface of bookshelf}\\ =\u20b9\text{\hspace{0.17em}\hspace{0.17em}}\frac{10}{100}\\ =\u20b9\text{\hspace{0.17em}}0.10{\text{per m}}^{\text{2}}\\ \mathrm{Total}\text{cost of polishing and painting the bookshelf}\\ =0.20\times 21700+0.10\times 19350\\ =4340+1935\\ =\u20b9\text{\hspace{0.17em}}6275\\ \text{Thus},\text{total expenses required for polishing and painting the}\\ \text{surface of the bookshelf is \u20b9 6275.}\end{array}$
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. Find the cost of paint required if silver paint costs 25 paise per cm^{2} and black paint costs 5 paise per cm^{2}.
Ans
$\begin{array}{l}\text{Diameter of wooden sphere}=21\text{\hspace{0.17em}}cm\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Radius of wooden sphere}=\frac{21}{2}\text{\hspace{0.17em}}cm\\ \text{\hspace{0.17em}Radius of each cylinder}=1.5\text{\hspace{0.17em}}cm\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Height}\text{of each cylinder}=7cm\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Surface area of sphere}=4\pi {r}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\times \frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=22\times 3\times 21\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1386\text{\hspace{0.17em}}c{m}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area of top of cylinder}=\pi {r}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}\times 1.5\times 1.5\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=7.07\text{\hspace{0.17em}}c{m}^{2}\\ \text{Surface area of sphere to be painted}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=13867.07\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1378.93\text{\hspace{0.17em}}c{m}^{2}\\ \text{Curved surface area of cylinder}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\times \frac{22}{7}\times 1.5\times 7\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=66\text{\hspace{0.17em}}c{m}^{2}\\ \text{Cost of silver painting on sphere}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{25}{100}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.\text{\hspace{0.17em}}0.25\\ \text{Cost of black painting on cylinder}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.\text{\hspace{0.17em}}\frac{5}{100}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.\text{\hspace{0.17em}}0.05\\ Cost\text{of painting of one sphere and cylinder}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.\text{\hspace{0.17em}}0.25\times 1378.93+\text{\hspace{0.17em}}Rs.\text{\hspace{0.17em}}0.05\times 66\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.(344.7325+3.3)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs\mathrm{.348.0325}\\ \text{Cost of painting of eight combinatin of sphere and cylinder}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8\times Rs\mathrm{.348.0325}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs\mathrm{.2784.26}\end{array}$
Q.3 The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Ans
\begin{array}{l}Let\text{diameter of sphere be 2r}\text{.}\\ \text{Curved surface area of sphere}=4\pi {r}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Decreased radius of sphere}=r25\%\text{of r}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=r\frac{25}{100}\times r\end{array} $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=r\frac{r}{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3\text{\hspace{0.17em}}r}{4}\\ \text{New curved surface area of sphere}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\pi {\left(\frac{3\text{\hspace{0.17em}}r}{4}\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{9}{4}\pi {r}^{2}\\ \text{Percentage of decreased curved surface area}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4\pi {r}^{2}\frac{9}{4}\pi {r}^{2}}{4\pi {r}^{2}}\times 100\%\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\frac{7}{4}\pi {r}^{2}}{4\pi {r}^{2}}\times 100\%\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{7}{16}\times 100\%\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=43.75\%\\ \text{Thus},\text{\hspace{0.17em}}\text{\hspace{0.17em}the 43}\text{.75\% curved surface area of sphere is decreased}\text{.}\end{array}$
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FAQs (Frequently Asked Questions)
1. Is Mathematics a complex subject?
Students might find that Mathematics is a complex and versatile academic discipline and they are eventually required to learn it. In addition, Mathematics is a significant, captivating, and amazing component of the reality of humans. With the aid of Mathematics, cuttingedge discoveries are made feasible in a number of important fields, including Engineering, Science, Commerce, and Technology. Students can learn various themes of Mathematics with the use of the scholarly materials available on the Extramarks website, such as the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.9. These resources can help students develop a healthy relationship with the subject.