# NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1

One of the key subjects for students in Class 9 is Mathematics. It is essential for students to have a foundational understanding of Mathematics at the school level in order to improve their ability to reason and think logically. Mathematical principles may be used by students to address difficulties in the real world. Every industry uses Mathematics in some capacity. A mathematical background can help students advance their careers in a variety of fields. Through the application of its concepts, students can pursue professions in Astronomy, Astrophysics, Statistics, Weather Forecasting, and other related disciplines. Mathematics is one of the core disciplines of a curriculum, it is studied in schools starting in the first grade. The foundational concepts of Mathematics are crucial for a student’s overall growth. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 may be used by students to get a thorough understanding of the mathematical principles discussed in the Class 9 Mathematics curriculum. Students in Class 9 may make use of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 to help them with their preparation for the senior secondary examination in Mathematics. It is essential that students pay attention to the mathematical principles taught in Class 9 since they are used as the foundation for the majority of competitive exams. Since most competitive examinations encapsulate the fundamentals of Class 9 Mathematics, understanding the topics taught in Class 9 Mathematics is essential to appearing in competitive examinations. If students plan to take competitive exams like the NEET, JEE Mains, JEE Advance, CAT, MAT, etc., they should pay particular attention to the mathematical concepts covered in Class 9 in order to do well on such exams. To accomplish the same, students in Class 9 can refer to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. Students can access the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1 from the Extramarks website.

**NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes **

To achieve high scores in Mathematics, students must consistently practice the NCERT solutions. In order for students to develop a firm command of their mathematical abilities, they must constantly revise and practice mathematical concepts. Students may practice the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 available on the Extramarks website. To properly understand the concepts discussed in Mathematics Chapter 13 Surface Areas and Volumes, students must be familiar with Class 9 Maths Chapter 13 Exercise 13.1. Students are suggested to use the Extramarks website’s NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. The necessary learning resources are available for download in PDF format from the Extramarks website. This feature has been made accessible by Extramarks to help students easily and quickly acquire the required study material. Students may also obtain the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 from the Extramarks website and Learning App. For their preparation for the senior secondary examination, Class 9 students may access the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 from Extramarks. For a variety of reasons, including issues with their internet connection, students occasionally find themselves unable to access educational resources. Due to this, Extramarks has provided the option for students to obtain the necessary learning resources in PDF format from the Extramarks website. Study materials are engineered by Extramarks with the requirements of all students in mind. The Extramarks website provides study and learning resources for all grades, from Class 1 to Class 12. Students can also access study materials for competitive exams via the Extramarks website and mobile application. With the assistance of Extramarks, students can participate in live online interactive lecture sessions from home. Furthermore, Extramarks offers interactive doubt-solving sessions for students to clarify their doubts effectively. These sessions help students study more effectively. Students in Class 9 can attend sessions to get answers to any questions they have about Math Class 9 Chapter 13 Exercise 13.1.

**Access NCERT Solutions for Class 9 Maths Chapter 13- Surface Areas and Volumes**

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 are available on the Extramarks website for Class 9 students. The Extramarks learning platform is well known for providing students with study materials and learning resources for a better educational experience. For all classes and boards, Extramarks offers a comprehensive overview of the syllabus. Extramarks is an e-learning portal that students from different classes and boards may access to prepare for exams. It is recommended that students in Class 9 students who are preparing for the senior secondary examination of Mathematics use the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. Students are recommended to consistently practice NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 to ensure thorough preparation. Students must thoroughly understand Class 9 Chapter 13 Maths Exercise 13.1 based on Surface Areas and Volumes. This may be accomplished by continuously practising the Extramarks-provided NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. For improved comprehension of the many mathematical topics covered in Chapter 13 Surface Areas and Volumes of the NCERT Mathematics textbook for Class 9, Exercise 13.1 Class 9 Maths solutions are helpful. The senior secondary examination of Mathematics may be prepared for using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. Students can improve their understanding of the mathematical concepts that are essential for performing well on the senior secondary Mathematics examination by continuously practising.

**NCERT Solutions For Class 9 Maths Chapter 13 Surface Area And Volume**

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For all classes, Extramarks offers credible and comprehensive NCERT solutions. The NCERT solutions to all exercise problems for all academic disciplines are available to students of all classes. Since the NCERT academic content is an inseparable part of the CBSE-prescribed academic curriculum, it is strongly advised that students who are taking the CBSE board examination study the NCERT solutions. It is advised that students in Class 9 use the NCERT solutions to assist themselves in their preparation. In order to be well-prepared for the senior secondary examination they will be taking, students may obtain the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 from the Extramarks website. Students may improve their understanding of the previously covered topics by regularly practising the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. While solving the exercise problems, it is advised that students use the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. For several reasons, doing so is advantageous. Proficient and qualified mentors who have prepared the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 are specialists in the subject arena of Mathematics. The solutions are designed in compliance with the CBSE Board’s specifications. The fact that the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 are divided and organised into steps and thoroughly explained assists students in their preparation. As they are simple to understand, they can be beneficial to students. Students can use the solutions as their primary text of reference for preparation since they are simple to understand. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 may be used by students to practice for the senior secondary Mathematics examination.

**NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1**

To score well in the senior secondary examination, students must have a thorough comprehension of each and every concept covered in the syllabus. Mathematical concepts can be challenging to grasp and apply. Students must be consistent in their practice of the exercise questions and answers if they want to be able to perform well in the designated examination. Students can practise solving the exercise problems using the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1. To achieve better grades in Mathematics, it is crucial to have a strong conceptual foundation. Students should be able to fully comprehend each concept, which may be achieved by regularly completing practice problems and solutions. In consideration of the senior secondary Mathematics examination, inculcating this routine may help students feel more confident. Students can get more confident and familiar with the curriculum by solving the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1. Moreover, it enhances students’ problem-solving skills. Students can learn to prioritise questions based on importance by regularly practising NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1, which will help them manage their time for the senior secondary Mathematics examination. For the objective of preparing for numerous competitive tests, Extramarks offers educational materials. Referring to Extramarks’ study resources is strongly advised for preparation for examinations like NEET, JEE, CUET, DUET, CAT, AIIMS, etc. It is advised for students who wish to take competitive examinations in the future to have a solid understanding of mathematical principles, considering they constitute a significant portion of those examinations. It is strongly recommended that students in Class 9 extensively study the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 in order to be well-prepared for competitive examinations. From the Extramarks website and mobile application, students can access adequate study materials for the preparation of competitive exams.

**NCERT Solutions for Class 9**

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**CBSE Study Materials for Class 9**

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**CBSE Study Materials**

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**Q.1 **A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:

(i) The area of the sheet required for making the box.

(ii) The cost of sheet for it, if a sheet measuring 1m^{2} costs ₹ 20.

**Ans**

$\begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Length of the box\hspace{0.33em}=\hspace{0.33em}1.5\hspace{0.17em}m}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Breadth of the box\hspace{0.33em}=\hspace{0.33em}1.25\hspace{0.17em}m}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Height of the box\hspace{0.33em}=\hspace{0.33em}65\hspace{0.17em}cm}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}0.65\hspace{0.17em}m}\\ \left(\text{i}\right)\text{\hspace{0.33em}Surface area of open box}\\ \text{=\hspace{0.33em}lb+2}\left(\text{b+l}\right)\text{h}\\ \text{=\hspace{0.33em}1.5\xd71.25+2}\left(\text{1.25+1.5}\right)\text{\xd70.65}\\ \text{=\hspace{0.33em}1.875+3.575}\\ {\text{=\hspace{0.33em}5.45\hspace{0.17em}m}}^{\text{2}}\\ \text{Thus, the area of the sheet required for making the box is}\\ {\text{5.45\hspace{0.17em}m}}^{\text{2}}\text{.}\\ \left(\text{ii}\right){\text{Cost of 1m}}^{\text{2}}\text{sheet\hspace{0.33em}=\hspace{0.33em}\u20b9\hspace{0.17em}20}\\ \text{Cost of sheet\hspace{0.33em}=\hspace{0.33em}\u20b9\hspace{0.17em}20\xd75.45\hspace{0.33em}=\hspace{0.33em}\u20b9109}\end{array}$

**Q.2 **The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m^{2}.

**Ans**

$\begin{array}{l}\text{The length of a room\hspace{0.33em}=\hspace{0.33em}5\hspace{0.17em}m}\\ \text{The breadth of a room\hspace{0.33em}=\hspace{0.33em}4\hspace{0.17em}m}\\ \text{The height of a room\hspace{0.33em}=\hspace{0.33em}3\hspace{0.17em}m}\\ \text{Surface area of 4 walls and ceiling\hspace{0.33em}=\hspace{0.33em}lb+2}\left(\text{b+l}\right)\text{h}\\ \text{\hspace{0.17em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}5\xd74+2}\left(\text{4+5}\right)\text{\xd73}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.17em}=\hspace{0.33em}20+54}\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}74\hspace{0.17em}m}}^{\text{2}}\\ {\text{Rate of whitewashing of 1 m}}^{\text{2}}\text{area\hspace{0.33em}=\hspace{0.33em}\u20b9\hspace{0.17em}7.50}\\ \text{Rate of whitewashing the room\hspace{0.33em}=\hspace{0.33em}\u20b9\hspace{0.17em}7.50\xd774}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}\u20b9 555}\\ \text{Thus, the cost of whitewashing the room is \u20b9 555.}\end{array}$

**Q.3** The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m^{2} is Rs 15000, fin the height of the hall.

**Ans**

$\begin{array}{l}\text{The perimeter of a rectangular hall\hspace{0.33em}=\hspace{0.33em}250\hspace{0.17em}m}\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Rate of painting the walls\hspace{0.33em}=\hspace{0.33em}\u20b910 per m}}^{\text{2}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}The cost of painting the 4 walls\hspace{0.33em}=\hspace{0.33em}\u20b915000}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Let height of hall\hspace{0.33em}=\hspace{0.33em}h\hspace{0.17em}m}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Area of 4 walls of the room\hspace{0.33em}=\hspace{0.33em}2}\left(\text{l+b}\right)\text{h}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}250\hspace{0.17em}h}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Cost of painting the four walls\hspace{0.33em}=\hspace{0.33em}10\xd7250\hspace{0.17em}h}\\ \text{\hspace{0.33em}\u20b9 15000\hspace{0.33em}=\hspace{0.33em}10\xd7250\hspace{0.17em}h}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}h\hspace{0.33em}=\hspace{0.33em}}\frac{\text{15000}}{\text{2500}}\text{\hspace{0.33em}=\hspace{0.33em}6\hspace{0.17em}m}\\ \text{Thus, the height of hall is 6 m.}\end{array}$

**Q.4** The paint in a certain container is sufficient to paint an area equal to 9.375 m^{2}. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

**Ans**

$\begin{array}{l}{\text{The area to paint by a certain container\hspace{0.33em}=\hspace{0.33em}9.375\hspace{0.17em}m}}^{\text{2}}\\ \text{Dimensions of given brick\hspace{0.17em}=\hspace{0.33em}22.5cm\xd710cm\xd77.5cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Surface area of a brick\hspace{0.33em}=\hspace{0.33em}2}\left(\begin{array}{l}\text{22.5cm\xd710cm+10cm\xd77.5cm}\\ \text{+7.5cm\xd722.5cm}\end{array}\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}2}\left(\text{225+75+168.25}\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}2}\left(\text{468.75}\right)\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}937.5\hspace{0.17em}cm}}^{\text{2}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\frac{\text{937.5}}{\text{10000}}{\text{\hspace{0.17em}m}}^{\text{2}}\\ \text{Number of bricks painted in given container\hspace{0.33em}=\hspace{0.33em}}\frac{\text{9.375}}{\left(\frac{\text{937.5}}{\text{10000}}\right)}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}100}\\ \text{Thus, 100 bricks can be painted out of this container.}\end{array}$

**Q.5** A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

(i) Which box has the greater lateral surface area and by how much?

(ii) Which box has the smaller total surface area and by how much?

**Ans**

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Edge of a cubical box}=10\text{\hspace{0.17em}}\mathrm{cm}\\ \text{Lateral\hspace{0.17em}}\mathrm{surface}\text{area of cubical box}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}=4{\left(10\right)}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=400\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length of cuboidal box}=\text{\hspace{0.17em}}12.5\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Breadth of cuboidal box}=\text{\hspace{0.17em}}10\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Height of cuboidal box}=\text{\hspace{0.17em}}8\text{\hspace{0.17em}}\mathrm{cm}\\ \text{Lateral\hspace{0.17em}}\mathrm{surface}\text{area of cuboidal box}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}2(\mathrm{l}+\mathrm{b})\mathrm{h}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2(12.5+10)\times 8\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=360\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{cubical box has the greater lateral surface area.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Greater area of cubical box}=400-360\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=40\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{(ii) Total surface area of cubical box}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6{\left(\mathrm{side}\right)}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6{\left(10\right)}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=600\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Total\hspace{0.17em}}\mathrm{surface}\text{area of cuboidal box}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}2(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2(12.5\times 10+10\times 8+8\times 12.5)\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2(125+80+100)\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left(305\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=610\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Total\hspace{0.17em}surface area of cuboidal box is greater than cubical box.}\\ \text{Greater surface area of cubical box}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=610-600\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=10\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\end{array}$

**Q.6 **A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

**Ans**

A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

**Q.7 **Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm^{2}, find the cost of cardboard required for supplying 250 boxes of each kind.

**Ans**

Total surface area of bigger box

= 2(lb + bh + hl)

= 2(25×20+ 20×5+ 5×25)

= 2 (725)

= 1450 cm^{2}

Required extra surface area = 5% of 1450 cm^{2}

=72.5 cm^{2}

Total surface area of smaller box

= 2(lb + bh + hl)

= 2(15×12+12×5+ 5×15)

= 2 (315)

= 630 cm^{2}

Required extra surface area = 5% of 630 cm^{2}

=31.5 cm^{2}

Total required cardboard for both types boxes

= 1450 + 72.5 + 630 + 31.5

= 2184 cm^{2}

Total required cardboard for both types 250 each boxes

= 250 x 2184 cm^{2}

= 546000 cm^{2}

Cost of 1000 cm^{2} cardboard = Rs. 4

Cost of required cardboard = Rs. (4/1000) x 546000

= Rs. 2184

**Q.8 **Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?

**Ans**

The length of shelter = 4 m

The breadth of shelter = 3 m

The height of shelter = 2.5 m

Total surface area of shelter of car

= lb + 2(b + l) × h

= 4 × 3 + 2(3+ 4)× 2.5 m^{2}

= 12 + 35

= 47 m^{2}

Thus, the required tarpaulin for shelter of car is 47 m^{2}.

## FAQs (Frequently Asked Questions)

### 1. How should students strategise while preparing for the Senior Secondary Examination of Mathematics of Class 9?

Students are expected to thoroughly go through each chapter and comprehend the concepts of all the topics and subtopics. To prepare the chapters properly, it is important to comprehend the fundamentals of each chapter. The most effective method for learning the fundamentals of Mathematics is problem-solving. Before tackling the harder questions, students should tackle the simpler ones first. Effective and efficient examination preparation is necessary for all students. Before starting their preparation for the Mathematics examination, students must become familiar with the CBSE curriculum. The syllabus is essential in order to plan a strategy for mastering the chapters of the Class 9 NCERT Mathematics textbook.

### 2. Why is using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 to prepare for the senior secondary examination advised for students of Class 9?

For their examination preparation, students of Class 9 are encouraged to use the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1, since they have been extensively prepared by subject specialists to help students comprehend the concepts of Mathematics better. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.1 on the Extramarks website and mobile application are extremely reliable and authentic resources. It is essential that students continue reviewing the chapters and solve as many exercise problems as they can in order to achieve higher scores on the Mathematics examination. Students must regularly solve sample papers and past years’ papers. While revising the chapters, it is advised to refer to the revision notes.