# NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.2

The area of a Trapezium, a generic Quadrilateral, and a Polygon are examined in-depth in the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 based on the chapter on Mensuration. The challenge with Class 8 Maths Exercise 11.2 is that there is not a single formula that can be applied in every case. Formulas are akin to broad concepts that must be customised in order to apply them to answer specific questions. There are several actions that may be taken, but developing a strong mathematical foundation is the key to solving these problems. There are 11 challenging problems encapsulated in the Class 8 Maths Chapter 11 Exercise 11.2. Students are therefore recommended to be well-prepared with a prior understanding of concepts to answer such queries.

Students should not be inhibited by this portion; instead, they should use it to further their understanding of the subject. If students run into any obstacles, they should refer to the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2. Students can also ask for assistance from their instructors and classmates to complete this activity successfully. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 can be found in downloadable PDF versions.

The formulae listed below are used in the problems in NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 Mensuration.

A Trapezium area equals half its height (sum of parallel sides). To get to this formula, a Trapezium is cut into a right-angled triangle and a rectangle. So, to obtain the generic formula for a Trapezium, students should add the areas of a Triangle and a Rectangle. The area of Polygons can also be found using this method. Before students can calculate the area of the provided Polygon, they must first fragment these figures into Polygons whose areas are already known.

The problems and questions from the exercise are thoroughly solved by the in-house subject matter experts of Extramarks in compliance with all CBSE standards. If a Class 8 student is thoroughly familiar with all of the concepts in the Mathematics textbook and appropriately educated about all of the exercises in it, they will easily score the highest on the annual exam. Students may easily understand the kinds of questions that may be a part of the test from this chapter, as well as the chapter’s weightage in terms of the total score, by using the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2, which allow them to be fully prepared for the annual exam. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 are credible learning resources available on the Extramarks Learning Portal and Extramarks Learning App.

**NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.2) Exercise 11.2 **

For the entireClass 8 Maths Chapter 11 Exercise 11.2, students can refer to the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2. To properly prepare for the Mathematics test, they can also access extra study materials, such as past years’ papers, revision notes, and study tips and techniques. Experienced educators of Mathematics have compiled substantial knowledge related to problem-solving in the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 in an easy-to-understand format. Furthermore, because the CBSE board strongly recommends studying from them, NCERT textbooks are a reliable resource for students to practise and succeed in the Mathematics examination. Mathematical questions will become easier to solve if students rely on the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2.

The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 contain several important formulae that students should be aware of. The formulae must not be mixed up since each geometric figure has a unique formula. Some important formulas are given below:

Area of a trapezium = ½ * Height * (sum of parallel sides)

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**Access NCERT solutions for Class 8 Chapter 11 – Mensuration**

The NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.2 pertaining to Mensuration Exercise 11.2, as well as similar solutions for additional exercises, are available online.Extramarks, India’s leading online learning platform, offers it. A PDF of the NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.2 based on Mensuration is available on the website.These solutions are quite beneficial while studying for tests. These solutions are developed by subject specialists with years of teaching experience and extensive subject knowledge.Students are recommended to get NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 Mensuration, for assistance with examination preparation.

Students should utilise the various learning resources provided on the Extramarks website. A proper combination of NCERT solutions, such as the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2, sample papers, past years’ papers, and so on, will help with Class 8 exam preparation.

**NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2**

Along with the concepts embedded in Class 8 Maths Exercise 11.2 Solution, this exercise has several important major concepts. To get the best scores in the class, it is essential to understand every idea in the textbooks and complete all the tasks that are given. Students ought to download the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 from the Extramarks website for better exam preparation. The NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.2 are available in high quality both offline and online.

Students will learn how to compute the areas of various geometric forms, such as Polygons, generic Quadrilaterals, and Trapezoids. These geometric forms are two-dimensional. In NCERT Class 8 Maths Chapter 11 Exercise 11.2, students will learn how to resolve problems based on these figures. Students must obtain the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 PDF from the Extramarks website to fully comprehend the numerous themes related to the exercise.

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There are 11 questions in the Class 8 Maths Chapter 11 Exercise 11.2. The themes “Area of a General Quadrilateral,” “Area of Special Quadrilaterals,” and “Area of Polygons” constitute the foundation for these problems. Students will learn a little bit about the topics by answering these questions, which will aid in the enhancement of their knowledge. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 are accessible as a PDF file on the Extramarks website and mobile application. These answers can assist them in removing any remaining uncertainties.

Many arithmetic operations are required in Class 8 Maths Exercise 11.2, which is based on the chapter on mensuration.Students must practise more if they are to do well in Class 8 Maths Chapter 11 Exercise 11.2 exams.First and foremost, students must study the theory and learn the formulae used in Exercise11.2. After that, students should solve each example and exercise question to gain a better understanding of the subject. Students are suggested to observe caution when performing the computations. After they have solved the problem, they could double-check their answer and method usingthe NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2.

**NCERT Solutions for Class 8**

The National Council of Educational Research and Training, or NCERT, has been charged with compiling and publishing academic textbooks and scholarly content for learners across the country. The NCERT textbooks are recognised by the CBSE and other State Boards and institutions as the mandatory reading for all exams. The NCERT textbooks are intended to provide students with the comprehensive knowledge they need for their whole mental development and to help them achieve exceptional achievements. The purpose of well-written NCERT textbooks is to provide students with the greatest education possible and encourage their capacity for critical and analytical thought. The NCERT books are standard-level materials, therefore it’s essential to have the proper tools and guidance to completely comprehend them.

For Class 8 students, Extramarks has provided NCERT Solutions Class 8 to help with their academics. A team of subject matter specialists created the NCERT solutions for Class 8, and the format is well-structured according to NCERT guidelines. Each solution is divided into steps so that students can easily understand it. The NCERT Class 8 textbook serves as the basis for the annual examination question papers. Thus, addressing issues with NCERT solutions for Class 8 helps students become ready for exams. Crucial study aid for exams are the NCERT solutions for Class 8. Writing has such an impact on the mind that when a student writes about a subject, they completely understand it. In addition to reviewing, students should set aside time for self-study. Students in Class 8 are required to take classes in subjects like Mathematics, which involves learning and proving a lot of formulae, Science, which involves learning and proving a lot of concepts from all three subsections, Physics, Chemistry and Biology and other subjects. The foundation for the curriculum of Class 10 is laid in classes 8 and 9. To prepare for the exam, students should start with the NCERT books before moving on to other texts.

Students can take a thorough look at the numerous formulae and sorts of questions that can arise on an exam with the aid of NCERT solutions for class 8 Mathematics. The various problem-solving methods encourage students to think creatively and approach problems with zeal.

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**Q.1 **The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

**Ans**

$\begin{array}{l}\text{Area of trapezium}=\frac{1}{2}\left(\text{Sum of parallel sides}\right)\times \left(\begin{array}{l}\text{Distances between}\\ \text{parallel sides}\end{array}\right)\\ \text{}=\left[\frac{1}{2}(1+1.2)(0.8)\right]{\mathrm{m}}^{2}\text{}\\ \text{}=0.88\text{}{\mathrm{m}}^{2}\end{array}$

**Q.2 **The area of a trapezium is 34 cm^{2} and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

**Ans**

$\begin{array}{l}\mathrm{It}\text{}\mathrm{is}\text{}\mathrm{given}\text{}\mathrm{that},\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{trapezium}=34\text{}{\mathrm{cm}}^{2}\\ \mathrm{and}\text{}\mathrm{height}=4\text{}\mathrm{cm}\\ \mathrm{Let}\text{}\mathrm{the}\text{}\mathrm{length}\text{}\mathrm{of}\text{}\mathrm{one}\text{}\mathrm{parallel}\text{}\mathrm{side}\text{}\mathrm{be}\text{}\mathrm{a}.\text{}\\ \mathrm{We}\text{}\mathrm{know}\text{}\mathrm{that},\\ \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{trapezium}=\frac{1}{2}\left(\mathrm{Sum}\text{}\mathrm{of}\text{}\mathrm{parallel}\text{}\mathrm{sides}\right)\times \left(\begin{array}{l}\mathrm{Distances}\text{}\mathrm{between}\\ \text{}\mathrm{parallel}\text{}\mathrm{sides}\end{array}\right)\end{array}$ $\begin{array}{l}34\text{}{\mathrm{cm}}^{2}=\frac{1}{2}(10\text{}\mathrm{cm}+\mathrm{a})\times 4\text{}\mathrm{cm}\\ \Rightarrow 34\text{}{\mathrm{cm}}^{2}=2(10\text{}\mathrm{cm}+\mathrm{a})\\ \Rightarrow 17\text{}\mathrm{cm}=10\text{}\mathrm{cm}+\mathrm{a}\\ \Rightarrow \mathrm{a}=17\text{}\mathrm{cm}-10\text{}\mathrm{cm}=7\text{}\mathrm{cm}\\ \\ \text{Thus},\text{the length of the other parallel side is 7 cm}.\end{array}$

**Q.3 **Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

**Ans**

$\begin{array}{l}\mathrm{Length}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{fence}\text{}\mathrm{of}\text{a}\mathrm{trapezium}\mathrm{shaped}\mathrm{field}\text{}\mathrm{ABCD}=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}\\ \Rightarrow 120\text{}\mathrm{m}=\mathrm{AB}+48\text{}\mathrm{m}+17\text{}\mathrm{m}+40\text{}\mathrm{m}\\ \Rightarrow \mathrm{AB}=120\text{}\mathrm{m}-105\text{}\mathrm{m}\\ =15\text{}\mathrm{m}\end{array}$ $\begin{array}{l}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{field}\text{}\mathrm{ABCD}=\frac{1}{2}(\mathrm{AD}+\mathrm{BC})\times \mathrm{AB}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\left[\frac{1}{2}(40+48)\times 15\right]{\mathrm{m}}^{2}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=\left(\frac{1}{2}\times 88\times 15\right){\mathrm{m}}^{2}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}=660\text{}{\mathrm{m}}^{2}\\ \\ \therefore \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{field}\text{}\mathrm{ABCD}=660\text{}{\mathrm{m}}^{2}\end{array}$

**Q.4 **The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

**Ans**

$\begin{array}{l}\text{It is given that},\\ \text{Length of the diagonal},\mathrm{d}=\text{24 m}\\ \text{Length of the perpendiculars},{\mathrm{h}}_{\text{1}}\text{and}{\mathrm{h}}_{\text{2}},\text{from the opposite vertices}\\ \text{to the diagonal are 8 m and 13 m}\\ \text{Area of the quadrilateral =}\frac{1}{2}\mathrm{d}\text{}\left({\mathrm{h}}_{1}+{\mathrm{h}}_{2}\right)\\ =\frac{1}{2}\times 24\text{}\times \left(13\text{}\mathrm{m}+8\text{m}\right)\\ =\frac{1}{2}\times 24\text{}\times 21\\ =252\text{}{\mathrm{m}}^{2}\\ \\ \text{Thus},{\text{the area of the field is 252 m}}^{\text{2}}.\end{array}$

**Q.5 **The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

**Ans**

\begin{array}{l}\text{Area of rhombus}=\frac{1}{2}\left(\text{Product of its diagonals}\right)\\ \text{Therefore},\text{area of the given rhombus}=\frac{1}{2}\left(\text{7}\text{.5\xd712}\right)\\ \text{}\text{}={\text{45 cm}}^{\text{2}}\end{array}

**Q.6 **Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

**Ans**

$\begin{array}{l}\text{Let the other diagonal of the rhombus be}x.\\ \text{Since rhombus is also a parallelogram}\text{.}\\ \text{Therefore, area of the given rhombus}=\text{Base}\times \text{Height}\\ =\text{5 cm}\times \text{4.8 cm}\\ ={\text{24 cm}}^{\text{2}}\\ \text{Also, area of rhombus}=\frac{1}{2}\left(\text{Product of its diagonals}\right)\\ \Rightarrow 24{\text{cm}}^{\text{2}}=\frac{1}{2}(\text{8 cm}\times \mathrm{x})\\ \Rightarrow \mathrm{x}=\frac{24\times 2}{8}\text{cm}=6\text{cm}\\ \\ \therefore \text{The length of the other diagonal of the rhombus is 6 cm and}\\ {\text{area of the given rhombus is 24 cm}}^{\text{2}}.\end{array}$

**Q.7 **The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m^{2} is ₹ 4.

**Ans**

$\begin{array}{l}\text{Area\hspace{0.33em}of\hspace{0.33em}rhombus\hspace{0.33em}=\hspace{0.33em}}\frac{\text{1}}{\text{2}}\text{\hspace{0.33em}}\left(\text{Product\hspace{0.33em}of\hspace{0.33em}its\hspace{0.33em}diagonals}\right)\\ \text{Area\hspace{0.33em}of\hspace{0.33em}each\hspace{0.33em}tile\hspace{0.33em}=\hspace{0.33em}}\left(\frac{\text{1}}{\text{2}}\text{\xd745\xd730}\right){\text{cm}}^{\text{2}}\\ {\text{= 675 cm}}^{\text{2}}\\ \text{Area\hspace{0.33em}of\hspace{0.33em}3000\hspace{0.33em}tiles\hspace{0.33em}=\hspace{0.33em}}\left(\text{675\xd73000}\right){\text{cm}}^{\text{2}}\\ {\text{\hspace{0.33em} \hspace{0.33em} \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}2025000 cm}}^{\text{2}}\\ {\text{\hspace{0.33em} \hspace{0.33em} \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}202.5 m}}^{\text{2}}\\ \therefore {\text{the\hspace{0.33em}total\hspace{0.33em}cost\hspace{0.33em}of\hspace{0.33em}polishing\hspace{0.33em}the\hspace{0.33em}floor,\hspace{0.33em}if\hspace{0.33em}the\hspace{0.33em}cost\hspace{0.33em}per\hspace{0.33em}m}}^{\text{2}}\text{\hspace{0.33em}is\hspace{0.33em}Rs\hspace{0.33em}4}\\ \begin{array}{l}=\u20b9\left(\text{4}\times \text{2}0\text{2}.\text{5}\right)\\ =\u20b9\text{81}0\end{array}\\ \text{Thus,\hspace{0.33em}the\hspace{0.33em}cost\hspace{0.33em}of\hspace{0.33em}polishing\hspace{0.33em}the\hspace{0.33em}floor\hspace{0.33em}is\hspace{0.33em}Rs\hspace{0.33em}810.}\end{array}$

**Q.8 **Mohan wants to buy a trapezium shaped field.

Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

**Ans**

\begin{array}{l}\text{Let the length of the field along the road be}\text{\hspace{0.17em}}l\text{\hspace{0.17em}}\text{m}.\text{}\\ \text{Hence},\text{the length of the field along the river will be 2}l\text{\hspace{0.17em}}\text{m}.\\ \text{Area of trapezium}=\frac{1}{2}\left(\text{Sum of parallel sides}\right)\text{}\left(\begin{array}{l}\text{Distance between}\\ \text{the parallel sides}\end{array}\right)\end{array} $\begin{array}{l}\Rightarrow 10500{\text{m}}^{\text{2}}=\frac{1}{2}(\mathrm{l}+2\mathrm{l})\text{}\left(100\text{m}\right)\\ \Rightarrow 3\mathrm{l}=\left(\frac{2\times 10500}{100}\right)\text{m}\\ \Rightarrow 3\mathrm{l}=\text{210 m}\\ \Rightarrow \mathrm{l}\text{}=\text{70 m}\\ \\ \\ \text{Thus},\text{length of the field along the river}=(\text{2}\times \text{7}0)\text{m}=\text{14}0\text{m}\\ \end{array}$

**Q.9** Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

**Ans**

Side of regular octagon = 5 cm

Area of trapezium ABCH = Area of trapezium DEFG

$\begin{array}{l}\text{Area of trapezium ABCH\hspace{0.33em}=\hspace{0.33em}}\left[\frac{\text{1}}{\text{2}}\text{\xd74\xd7}\left(\text{11+5}\right)\right]{\text{m}}^{\text{2}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\left[\frac{\text{1}}{\text{2}}\text{\xd74\xd716}\right]{\text{m}}^{\text{2}}\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}\hspace{0.33em}32\hspace{0.17em}m}}^{\text{2}}\end{array}$

Area of rectangle HGDC = 11 × 5 = 55 m^{2}

Area of octagon = Area of trapezium ABCH + Area of trapezium DEFG + Area of rectangle HGDC

= 32 m^{2 }+ 32 m^{2 }+ 55 m^{2} = 119 m^{2}

Therefore, the area of octagonal surface is 119 m^{2}

**Q.10 **There is a pentagonal shaped park as shown in the figure.

For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

**Ans**

__Jyoti way of calculating area:__

\begin{array}{l}\text{Area of pentagon}=\text{2}\left(\text{Area of trapezium ABCF}\right)\\ =\left[2\times \frac{1}{2}\times (15+30)\left(\frac{15}{2}\right)\right]{\text{m}}^{2}\\ =337.5{\text{m}}^{2}\end{array}

__Kavita’s way of calculating area__:

\begin{array}{l}\text{Area of pentagon ABCDE}=\text{Area of \Delta ABE}+\text{Area of square BCDE}\\ \text{}=\left[\frac{1}{2}\times 15\times \left(30-15\right)+{\left(15\right)}^{2}\right]{\text{m}}^{2}\\ \text{}=\left[\frac{1}{2}\times 15\times 15+225\right]{\text{m}}^{2}\\ \text{}=337.5{\text{m}}^{2}\end{array}

No, there is no other way to find its area.

**Q.11 **Diagram of the adjacent picture frame has outer dimensions 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

**Ans**

It is given that the width of each section is same.

Therefore,

LM = BM = CN = NE = OF = OH = PK = PI

CH = CN + NO+ OH

28 = 2CN+20

2CN = 28 – 20

2CN = 8

CN = 4 cm

Hence, LM = BM = NE = OF = OH = PK = PI = 4 cm

$\begin{array}{l}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{section}\text{}\mathrm{NDGO}\text{}=\text{}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{section}\text{}\mathrm{AMPJ}\\ =\left[\frac{1}{2}(20+28)\left(4\right)\right]{\mathrm{cm}}^{2}\\ =96\text{}{\mathrm{cm}}^{2}\\ \therefore \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{section}\text{}\mathrm{NDGO}\text{}=\text{}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{section}\text{}\mathrm{AMPJ}=96\text{}{\mathrm{cm}}^{2}\\ \\ \mathrm{Also},\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{section}\text{}\mathrm{AMND}=\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{section}\text{}\mathrm{PJGO}\\ =\left[\frac{1}{2}(16+24)\left(4\right)\right]\\ =80\mathrm{}{\mathrm{cm}}^{2}\\ \therefore \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{section}\text{}\mathrm{AMND}=\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{section}\text{}\mathrm{PJGO}=80\mathrm{}{\mathrm{cm}}^{2}\end{array}$

##### FAQs (Frequently Asked Questions)

## 1. What significance do the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 have?

In addition to the in-depth reading of the NCERT textbooks, the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 provide a complete comprehension of Mensuration and related concepts. Class 8 Maths Exercise 11.2 Solution follows a logical framework that makes it easy to understand complex ideas. With the help of the various examples and practice questions provided along with the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2, students can rapidly learn the best approach for performing well on annual examinations. Due to their exceptional quality, NCERT solutions are a very reliable tool for learning and reviewing for exams.

## 2. Where can students find the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2?

The NCERT Class 8 Maths Chapter 11 Exercise 11.2 and Mensuration for Class 8 Mathematics are available on the Extramarks website.It is possible to access the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 online through the Extramarks website. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2, which include step-by-step solutions to the exercise questions, are available in PDF format on Extramarks. The NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2 are based on the most recent NCERT guidelines and standards.

## 3. How many chapters are there in the NCERT textbook of Mathematics for class 8?

There are 16 chapters in total in the NCERT textbook of Class 8 for Mathematics. These are as follows-

Chapter 1 – Rational Numbers

Chapter 2 – Linear Equations in One Variable

Chapter 3 – Understanding Quadrilaterals

Chapter 4 – Practical Geometry

Chapter 5 – Data Handling

Chapter 6 – Squares and Square Roots

Chapter 7 – Cubes and Cube Roots

Chapter 8 – Comparing Quantities

Chapter 9 – Algebraic Expressions and Identities

Chapter 10 – Visualising Solid Shapes

Chapter 11 – Mensuration

Chapter 12 – Exponents and Powers

Chapter 13 – Direct and Inverse Proportions

Chapter 14 – Factorisation

Chapter 15 – Introduction to Graphs

Chapter 16 – Playing with Numbers

## 4. What is the best way to cover NCERT Class 8 Maths Chapter 11 exercise 11.2?

The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 are the simplest approach to begin preparing for Class 8 Maths Chapter 11 Exercise 11.2. The learner will save time by using the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.2 resources. In addition, students should develop the practise of taking notes so that they may be reviewed throughout the preparation process. Making notes and consulting the NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2 can help students remember what they have learned.