# NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.3

The Central Board of Secondary Education CBSE is one of the biggest and most highly reputed educational boards operating in India. There is a plethora of schools affiliated with the CBSE board, which are mostly present in all the leading cities in IndiaCBSE follows an extremely objective approach when it comes to its education pattern. CBSE values a study environment which ensures that students understand the topics and ideas that they are learning. CBSE trains its students to have a tendency, whenever they are studying something new and learning new ideas, to do so with great depth and a formidable base. Students must learn how to delve deeply into topics and ideasCBSE trains its students to learn to sift through the redundant information to find the important information. The questions that are asked in the exams follow an extremely objective pattern. An objective pattern entails questions which are mainly short and direct. The questions that are generally asked in the exams are short, but they are deeply entrenched in the fundamentals of the topic. If students are not familiar with the text and have not thoroughly understood the concepts, they will fail to answer the questions correctly. The learning process in a CBSE curriculum is steep and requires an extensive amount of hard work and dedication. Students have to be consistent with their studying. The Teachers who are a part of Extramarks have observed all the struggles that students generally go through, and therefore they have shared various tips and tricks that have seemed to work.

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The National Council of Educational Research and Training, NCERT is a centrally controlled organisation which was founded in 1961. NCERT was founded by the Indian government so that India could have an autonomous branch in its education department which would make decisions which are complicated and related to the educational advancement of the country. NCERT is comprised of highly qualified individuals who are qualified in making executive decisions in matters regarding the academic development of the country and its prowess. NCERT and CBSE share an extremely symbiotic relationship, and they make decisions together. NCERT puts out their own sets of rules and regulations, and students must strictly adhere to these rules if they wish to score higher marks in the exams. NCERT even curates their syllabus, and they demand all the schools affiliated with the CBSE board strictly follow the syllabus. NCERT has its own publishing department, which publishes books that stringently follow the syllabus as well as all the rules and regulations. Teachers at Extramarks have shared the importance for students to make their NCERT textbook their primary textbook. Students often get overwhelmed with so much information, and the stature of the CBSE board often does not help. Teachers have reassured students that there is nothing for them to find daunting or be scared of.

Mathematics is one of the most important and challenging subjects that a Class 8 student studies as part of their curriculum. Every student in Class 8 following the CBSE curriculum has Mathematics as a subject. Students who have been promoted to Class 8 are in their final year of middle school. Therefore, CBSE tries to prepare its students for the advanced level of education practised at the high school level. Generally, students are a little avoidant of Mathematics and the primary reason among students for this is that the subject is difficult, although teachers have noted that it is not the level of difficulty that students are often scared of. Students are generally more worried about the process that must be followed if they want to do well in Mathematics. Students must be precise, and they must be very regular with their practice. Given the stature of the syllabus and the scale of the curriculum, students often fail to provide the necessary attention to this subject. Teachers have reassured students that there is nothing tobe scared of regarding the subject. If students incorporate the little changes that teachers ask them to make, then teachers can rest assured that students will not be dealing with the problems that they generally deal with regarding the subject.

Class 8 Maths Exercise 11.3 is an important exercise and students must be very careful while learning about this chapter. Students make a lot of silly errors while solving Class 8 Maths Chapter 11 Exercise 11.3 and therefore the solutions help students with it.

**NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.3) Exercise 11.3**

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**Access NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.3**

Chapter 11 in the Class 8 curriculum is Mensuration. Students learn about the different shapes and properties of basic 2-d and 3-d objects. Students learn about the formulas used to calculate the Volume, Surface Area, Perimeter and Area of these shapes. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 provide solutions to this chapter with great clarity.

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**Exercise 11.3**

Teachers at Extramarks have years of experience teaching CBSE students and given their experience, they have observed all the struggles that these students go through. Therefore the teachers have come together to make the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 is a collection of all the NCERT solutions for every unsolved problem in the NCERT textbook. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 follow the exact chronology and the same question numbers as given in the NCERT textbook. This is done to ensure that students have no trouble following the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 while they are solving problems.

The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 provide extremely direct and easy-to-understand solutions to every NCERT question. The solutions are provided in the most lucid language possible. The solutions are written in a very comprehensible way, and they are conveniently comprehensive as well.

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The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 are compiled by qualified professionals. Therefore students who are using the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 are always under expert guidance. The solutions are highly reliable and free of inaccuracies. Therefore, students can trust these NCERT solutions. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 are well detailed, which ensures that students go to the depths of the topic. The NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.3 is one of the most effective and powerful study resources available to any Class 8 CBSE candidate.These solutions greatly help students get prepared for their exams. The solutions are easy to understand, and students seldom retain their doubts after they have referred to the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3. Chapter-wise NCERT Solutions for Class 8 Maths

Extramarks releases various resources, like the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3; they are available for almost all the classes and all the subjects included in the curriculum. Students must go through the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 when they come across a doubt. TThe manner in which solutions are presented in the NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.3, aids students immediately.It is unusual for students to have questions after consulting the NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.3.

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**Class 8 Maths Chapter 11 Includes:**

The third Exercise of Chapter 11 teaches students about the formulas and properties of 3-D shapes like cubes, Cuboids, Cylinders etc. the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 helps students understand these concepts very easily.

**NCERT Solutions for Class 8**

Click on the link below to access the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 in English and Hindi. When students refer to the NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.3 on a regular basis, they become acquainted with the pattern of the exam’s question paper.The solutions inform students of the correct way of answering the CBSE questions,which would ensure maximum marks. When using the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.3 students get a clear understanding of the weightage that each chapter carries, and they also learn how to efficiently manage their time. Students can access a variety of other learning resources from the Extramarks website. Past years’ papers, sample question papers, revision notes, andlive doubt-solving sessions are some of the frequently used learning tools on the Extramarks website.

**Q.1 **There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to

**Ans**

$\begin{array}{l}\text{We know that},\text{Total surface area of the cuboid}=\text{2}(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})\\ \\ \text{Total surface area of cuboid}=\left[\text{2}\{\left(\text{6}0\right)\left(\text{4}0\right)+\left(\text{4}0\right)\left(\text{5}0\right)+\left(\text{5}0\right)\left(\text{6}0\right)\}\right]{\text{cm}}^{\text{2}}\\ =\left[\text{2}(\text{24}00+\text{2}000+\text{3}000)\right]{\text{cm}}^{\text{2}}\\ =\left[\text{2}(\text{24}00+\text{2}000+\text{3}000)\right](\text{2}\times \text{74}00){\text{cm}}^{\text{2}}\\ =\left[\text{2}(\text{24}00+\text{2}000+\text{3}000)\right]\text{148}00{\text{cm}}^{\text{2}}\\ \\ \text{Total surface area of the cube}=\text{6}{\left(\mathrm{l}\right)}^{\text{2}}\\ =\text{6}{\left(\text{5}0\text{cm}\right)}^{\text{2}}\\ =\text{15}000{\text{cm}}^{\text{2}}\\ \\ \text{Since the total surface area of cuboid is less than}\\ \text{the total surface area of cube,therefore, the cuboidal box will}\\ \text{require lesser amount of material}.\end{array}$

**Q.2 **A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

**Ans**

$\begin{array}{l}\mathrm{A}\text{}\mathrm{suitcase}\text{}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{shape}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{cuboid}.\\ \therefore \mathrm{Total}\text{}\mathrm{surface}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{cuboid}=\text{}2(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})\\ \mathrm{Hence},\\ \mathrm{Total}\text{}\mathrm{surface}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{suitcase}=2[\left(80\right)\left(48\right)+\left(48\right)\left(24\right)+\left(24\right)\left(80\right)]\text{}{\mathrm{cm}}^{2}\\ =2[3840\text{}+\text{}1152\text{}+\text{}1920]\text{}{\mathrm{cm}}^{2}\\ =13824\text{}{\mathrm{cm}}^{2}\\ \therefore \mathrm{Total}\text{}\mathrm{surface}\text{}\mathrm{area}\text{}\mathrm{of}\text{}100\text{}\mathrm{suitcases}=(13824\text{}\times \text{}100){\mathrm{cm}}^{2}\\ =1382400\text{}{\mathrm{cm}}^{2}\\ \\ \mathrm{Required}\text{}\mathrm{tarpaulin}=\mathrm{Length}\times \mathrm{Breadth}\\ \Rightarrow \text{}1382400\text{}{\mathrm{cm}}^{2}\text{}=\mathrm{Length}\times 96\text{}\mathrm{cm}\\ \mathrm{Length}=\left(\frac{1382400}{96}\right)\mathrm{cm}=14400\text{}\mathrm{cm}=144\text{}\mathrm{m}\\ \mathrm{Thus},\text{}144\text{}\mathrm{m}\text{}\mathrm{of}\text{}\mathrm{tarpaulin}\text{}\mathrm{is}\text{}\mathrm{required}\text{}\mathrm{to}\text{}\mathrm{cover}\text{}100\text{}\mathrm{suitcases}.\end{array}$

**Q.3** Find the side of a cube whose surface area is 600 cm^{2}.

**Ans**

$\begin{array}{l}\mathrm{Given}\text{}\mathrm{that},\text{}\mathrm{surface}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{cube}=600\text{}{\mathrm{cm}}^{2}\\ \mathrm{Let}\text{}\mathrm{the}\text{}\mathrm{length}\text{}\mathrm{of}\text{}\mathrm{each}\text{}\mathrm{side}\text{}\mathrm{of}\text{}\mathrm{cube}\text{}\mathrm{be}\text{\u2018}\mathrm{x}\u2018.\\ \mathrm{Surface}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{cube}=6\text{}{\left(\mathrm{Side}\right)}^{2}\end{array}$ $\begin{array}{l}\Rightarrow 600\text{}{\mathrm{cm}}^{2}=6{\mathrm{x}}^{2}\\ \Rightarrow {\mathrm{x}}^{2}=100\text{}{\mathrm{cm}}^{2}\\ \Rightarrow \mathrm{x}=10\text{}\mathrm{cm}\\ \\ \therefore \text{}\mathrm{The}\text{}\mathrm{side}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{cube}\text{}\mathrm{is}\text{}10\text{}\mathrm{cm}.\end{array}$

**Q.4 **Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

**Ans**

$\begin{array}{l}\text{Length of the cabinet}=\text{2 m}\\ \text{Breadth of the cabinet}=\text{1 m}\\ \text{Height of the cabinet}=\text{1}.\text{5 m}\end{array}$ $\begin{array}{l}\text{Area of the cabinet that was painted}=\text{2}\mathrm{h}(\mathrm{l}+\mathrm{b})+\mathrm{lb}\\ =\text{}[\text{2}\times \text{1}.\text{5}\times (\text{2}+\text{1})+\left(\text{2}\right)\left(\text{1}\right)]{\text{m}}^{\text{2}}\\ =\text{}[\text{3}\left(\text{3}\right)+\text{2}]{\text{m}}^{\text{2}}\\ =\text{}(\text{9}+\text{2}){\text{m}}^{\text{2}}\\ ={\text{11 m}}^{\text{2}}\\ \\ \therefore {\text{Area of the cabinet that was painted is 11 m}}^{\text{2}}.\end{array}$

**Q.5 **Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m^{2} of area is painted. How many cans of paint will she need to paint the room?

**Ans**

$\begin{array}{l}\text{Given that},\\ \text{Length=15 m}\\ \text{breadth=1}0\text{m}\\ \text{height=7 m}\\ \\ \text{Area of the hall to be painted}=\left(\begin{array}{l}\text{Area of}\\ \text{the walls}\end{array}\right)+\left(\begin{array}{l}\text{Area of}\\ \text{the ceiling}\end{array}\right)\end{array}$ $\begin{array}{l}=\text{2}\mathrm{h}(\mathrm{l}+\mathrm{b})+\mathrm{lb}\\ =\left[\text{2}\left(\text{7}\right)(\text{15}+\text{1}0)+\text{15}\times \text{1}0\right]{\text{m}}^{\text{2}}\\ =\left[\text{14}\left(\text{25}\right)+\text{15}0\right]{\text{m}}^{\text{2}}\\ =\text{5}00{\text{m}}^{\text{2}}\\ \\ \text{It is given that 1}00{\text{m}}^{\text{2}}\text{area can be painted from each can}.\\ \\ \therefore \text{Number of cans required to paint an area of 5}00{\text{m}}^{\text{2}}=\frac{500}{100}=5\\ \\ \text{Hence},\text{5 cans are required to paint the walls and the ceiling}\\ \text{of the cuboidal hall}.\end{array}$

**Q.6 **Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?

**Ans**

$\begin{array}{l}\text{Both the figures have the same heights and the difference}\\ \text{between the two figures is that one is a cylinder}\\ \text{and the other is a cube}.\end{array}$ $\begin{array}{l}\mathrm{Now},\\ \text{Lateral surface area of the cube}=\text{4}{\mathrm{l}}^{\text{2}}=\text{4}{\left(\text{7 cm}\right)}^{\text{2}}={\text{196 cm}}^{\text{2}}\\ \text{Lateral surface area of the cylinder}=\text{2}\mathrm{\pi rh}\mathrm{sq}.\mathrm{units}\\ =(2\times \frac{22}{7}\text{\xd7}\frac{7}{2}\text{\xd77}){\text{cm}}^{\text{2}}\\ ={\text{154 cm}}^{\text{2}}\\ \\ \text{Hence},\text{the cube has larger lateral surface area}.\end{array}$

**Q.7 **A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

**Ans**

$\begin{array}{l}\text{Total surface area of cylinder}=\text{2}\mathrm{\pi r}(\mathrm{r}+\mathrm{h})\mathrm{sq}.\mathrm{units}\\ =\left[2\times \frac{22}{7}\times 7(7+3)\right]{\mathrm{m}}^{2}\\ =\text{44}0{\text{m}}^{\text{2}}\\ \\ \text{Thus},\text{44}0{\text{m}}^{\text{2}}\text{sheet of metal is required}.\end{array}$

**Q.8 **The lateral surface area of a hollow cylinder is 4224 cm^{2}. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet.

**Ans**

$\begin{array}{l}\text{Surface area of hollow cylinder}=\text{Area of rectangular sheet}\\ \Rightarrow {\text{4224 cm}}^{\text{2}}=\text{Length}\times 33\mathrm{cm}\\ \Rightarrow \text{Length =}\frac{4224}{33}=128\text{}\mathrm{cm}\\ \\ \therefore \text{The length of the rectangular sheet is 128 cm}.\\ \\ \text{Perimeter of the rectangular sheet}=\text{2}(\text{Length}+\text{Width})\\ =\left[\text{2}(\text{128}+\text{33})\right]\text{cm}\\ =(\text{2}\times \text{161})\text{cm}\\ =\text{322 cm}\\ \\ \therefore \text{Perimeter of the rectangular sheet}=322\text{}\mathrm{cm}.\end{array}$

**Q.9** A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

**Ans**

$\begin{array}{l}\text{Given}\\ \text{Diameter= 84 cm}\Rightarrow \text{r = 42 cm=}\frac{\text{42}}{100}\text{m}\\ \text{Length=1 m}\end{array}$ $\begin{array}{l}\text{In one revolution},\text{the roller will cover an area equal to its}\\ \text{lateral surface area}.\\ \end{array}$ \begin{array}{l}\text{}\end{array} $\begin{array}{l}\text{Lateral surface area of cylinder}=\text{2}\mathrm{\pi rh}\\ =2\times \frac{22}{7}\times 42\text{}\mathrm{cm}\times 1\text{}\mathrm{m}\\ =2\times \frac{22}{7}\times \frac{42\text{}}{100}\mathrm{m}\times 1\text{}\mathrm{m}\\ =\frac{264}{100}{\text{m}}^{\text{2}}\\ \\ \therefore \text{In 75}0\text{revolutions},\text{area of the road covered}=750\times \frac{264}{100}{\text{m}}^{\text{2}}\\ \text{}=\text{198}0{\text{m}}^{\text{2}}\end{array}$

**Q.10 **A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

**Ans**

$\begin{array}{l}\mathrm{Height}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{label}=20\text{}\mathrm{cm}-2\text{}\mathrm{cm}-2\text{}\mathrm{cm}=16\text{}\mathrm{cm}\\ \mathrm{Diameter}=14\text{}\mathrm{cm}\\ \Rightarrow \mathrm{Radius}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{label}=\frac{14}{2}\text{}\mathrm{cm}=\text{}7\text{}\mathrm{cm}\\ \\ \mathrm{Label}\text{}\mathrm{is}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{form}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{cylinder}\text{}\mathrm{having}\text{}\mathrm{its}\text{}\\ \mathrm{radius}\text{}\mathrm{and}\text{}\mathrm{height}\text{}\mathrm{as}\text{}7\text{}\mathrm{cm}\text{}\mathrm{and}\text{}16\text{}\mathrm{cm}.\end{array}$ $\begin{array}{l}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{label}=2\mathrm{\pi rh}\\ =(2\times \frac{22}{7}\times 7\times 16){\mathrm{cm}}^{2}\\ =704\text{}{\mathrm{cm}}^{2}\\ \\ \therefore \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{label}=704\text{}{\mathrm{cm}}^{2}\end{array}$

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