NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula (Ex 12.2) Exercise 12.2

Mathematics is one of the most basic skills that every student must learn. It is taught as a subject in most schools. Students should focus on practising questions related to all the topics included in the Mathematics syllabus. Practising Mathematics problems helpsenhance the analytical thinking ability and problem-solving skills of students. They are able to make better decisions in their lives with the knowledge of Mathematical principles.

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It is important for students to practice questions given in the NCERT textbook of Mathematics to perform well in the examinations. The exercises given in the NCERT textbook are very crucial from the examination perspective. Students may have difficulties solving questions given in the exercises. In that case, they can access NCERT solutions from the Extramarks website and mobile application. All exercises in each chapter of Class 9 Mathematics can be practised using the NCERT solutions. To practice Exercise 12.2 of Chapter 12, Class 9 students can use the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2.

The main topic of Chapter 12 is Heron’s Formula. It is one of the important chapters from an examination standpoint. All the theoretical portions of Chapter 12 should be well learned by students. It is crucial for students to review the theorems, formulae, definitions, derivations, etc. of Chapter 12 to be able to solve the exercises in the chapter. There are two exercises in Chapter 12, and both of them should be practised consistently to prepare well for the upcoming examination in Mathematics. Exercise 12.2 is crucial from the examination point of view. Students should review the theories of the topics before beginning to solve the exercise questions. If students have doubts about solving Exercise 12.2, they can make use of the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2. Some of the questions in Exercise 12.2 can seem challenging for students. They can learn to break down complex problems into small and simpler steps with the assistance of the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2.

Regular practice is the key to performing well in the Mathematics examination. It is necessary to develop a habit of solving questions regularly. Practising questions with the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2 is beneficial for developing a habit of solving questions. There are a variety of questions asked in the Class 9 Maths Chapter 12 Exercise 12.2, students must learn to solve each type of question to score well in the Mathematics examination. The Class 9 Maths Exercise 12.2 Solution will assist students to practice all the types of questions asked in the Maths Class 9 Chapter 12 Exercise 12.2. All the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2 are helpful in boosting the confidence of students as well. They get motivated to solve more questions and their preparation also increases. The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2 are beneficial for learning time management skills for students. They will be able to solve questions within the allotted time when they get used to solving questions regularly.

There are solved examples given after each topic in Chapter 12. Students are supposed to go through all the solved examples for a better understanding of the ways to solve questions. Solved examples are intended to show students how to use formulas and find answers to questions.  Students are required to pay attention to solved examples from Chapter 12 as well before starting to solve exercises. Students will find it helpful to solve exercise problems if they refer to the solved examples. The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2 are beneficial for removing doubts of students related to Exercise 12.2. It is crucial to understand the requirements of the questions asked before starting to solve them. The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, will also assist students in understanding the requirements of the questions asked in Exercise 12.2 Class 9 Maths.

Extramarks is a learning platform dedicated to providing the best quality learning materials for students of all classes. It is a very trustworthy online learning portal that has been trusted by more than 10,000 leading Indian schools since 2007. Students of CBSE, ICSE, and other major state boards can score well in their examinations with the assistance of Extramarks. Students can also take assistance from Extramarks to obtain higher marks in competitive examinations like JEE, CUET, NEET, etc. All these examinations are conducted every year and are very crucial for getting admissions into the top colleges in India. Extramarks assists students to score well in the Olympiad examinations as well. Students can access a variety of practice materials and mock tests to assess their preparation level. All the study materials provided by Extramarks are regularly updated and verified. Extramarks also helps students of Class 9 to score well in the Mathematics examinations by providing the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2.

In addition to practising exercises, it is essential to practice past years’ papers and sample papers of Mathematics from time to time. Students are advised to access past years’ papers and sample papers from Extramarks and practice them regularly. Past years’ papers are helpful in practising the types of questions that can appear in the Mathematics examination. Students will also get familiar with the marking scheme and paper pattern of Mathematics when they go through the past years’ papers of Mathematics. Class 9 students can also access detailed solutions for questions given in the past years’ papers from Extramarks.

The live lectures of Extramarks are very significant for the preparation of examinations. Class 9 students can make use of the live lectures to learn the theories of Chapter 12. All the topics are explained well by the best teachers of Mathematics. Extramarks teachers are readily available to answer students’ questions about the topics. All the definitions, formulae, etc. of Chapter 12 can be understood well with the guidance and support of Extramarks’ in-house teachers of Mathematics.

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula (Ex 12.2) Exercise 12.2

Students must access the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, from the Extramarks website and mobile application. All the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2 are available in a well-structured format and can be easily referenced. Students can trust the authenticity of the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, as they are prepared by expert Mathematics teachers. It is crucial to manage time efficiently while practising questions because the time allotted for solving question papers of Mathematics is fixed. Students are advised to practice questions under timed conditions to improve their time management skills. When they practise questions with the NCERT Solutions for Class 9 Maths Chapter 12 Exercise 12.2, they will improve their time management skills.

Access NCERT Solutions for Class 9 Mathematics Chapter 12 – Heron’s Formula

The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2 are accessible in PDF format on the Extramarks website and mobile application. Students can utilise them in offline mode as well.

Exercise 12.2

Exercise 12.2 has 9 questions based on the topic of the Application of Heron’s Formula in Finding Areas of Quadrilaterals. All the questions should be practised well by students in order to score higher marks in the Mathematics examination.

Class 9th Maths NCERT Solutions Chapter 12.2

The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, are available in PDF format on the Extramarks learning platform. Students can easily solve Exercise 12.2 with these solutions. The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, are helpful in solving any question that may appear from Exercise 12.2 in the Mathematics examination.

NCERT Solutions for Class 9 Maths 

All the exercises of each chapter in Class 9 Mathematics can be solved with the help of the NCERT solutions. Students can access the NCERT solutions for all the chapters of Class 9 Mathematics from Extramarks. These solutions are helpful in enhancing the preparation level for the Mathematics examination.

NCERT Class 9 Maths Chapter 12 Exercise 12.2 Question 1

Question 1 given in Exercise 12.2 is based on the application of Heron’s Formula in Finding Areas of Quadrilaterals. It is important to understand the questions before starting to solve them. Students can make use of the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2 to solve question 1 of Exercise 12.2.

Class 9 Maths Chapter 12 Exercise 12.2 Question 2

Students having difficulty in solving question 2 of Exercise 12.2 can take the assistance of the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2. It is important to revise Heron’s formula to solve question 2.

Exercise 12.2 Class 9 Question 3

Question 3 of Exercise 12.2 requires students to find the area of the paper used to make a shape. To solve this question, students should understand the question well along with the given figure. Question 2 can be solved with the aid of the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2.

NCERT Maths Class 9 Chapter 12 Exercise 12.2 Question 4

Students are supposed to find the height of the parallelogram in Question 2. To find the appropriate solution to question 4 of Exercise 12.2, it is necessary to access the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2.

Maths NCERT Solutions Class 9 Chapter 12 Exercise 12.2 Question 5

Class 9 students must refer to the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2 to solve question 5 of Exercise 12.2.

Class 9 Maths NCERT Solutions Chapter 12 Exercise 12.2 Question 6

It is necessary to focus on the figure given in question 6 to solve it. Understanding the question will enable students to come up with a solution. The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, should be referred to while solving question 6 of Exercise 12.2.

Class 9 Maths Chapter 12 Exercise 12.2 Solutions Question 7

Students should read question 7 carefully to understand its requirements. They are supposed to refer to the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, to get detailed solutions to question 7 in Exercise 12.2.

Ex 12.2 Class 9 Question 8

Class 9 students are required to find the cost of polishing the tiles on the floor as it is given in the figure with question 8 in Exercise 12.2. Students can solve this question by referring to the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2.

Maths NCERT Solutions Class 9 Chapter 12 Exercise 12.2 Question 9

Question 9 of Exercise 12.2 can be solved easily with the assistance of the NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2. Students are supposed to have clarity of the question before solving it.

Key Takeaways of NCERT Solutions Class 9 Maths Chapter 12 Exercise 12.2

The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, are very significant for scoring well in the Mathematics examination. All the solutions are easy to understand. The NCERT Solutions For Class 9 Maths Chapter 12 Exercise 12.2, are accessible on the Extramarks learning platform in PDF format.

NCERT Solutions for Class 9

All the subjects in Class 9 can be understood well by solving exercises with the NCERT solutions. Students can access them on Extramarks. They are prepared according to the latest CBSE syllabus for each subject. Students can download the NCERT solutions for Mathematics, English, Hindi, etc. from the Extramarks learning platform.

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Q.1 A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Ans

In​ ΔBCD, ∠BCD=90°So, Pythagoras theorem   BD=BC2+CD2      =(12)2+(5)2       =144+25       =169       =13 cmar(ΔBCD)      =12×BC×DC      =12×12×5      =30cm2Area ΔABD, by Heron’s formula     Δ=s(s−a)(s−b)(s−c)      s=a+b+c2       =9+8+132       =302=15cm   Δ=15(15−9)(15−8)(15−13)       =15×6×7×2      =635 cm      =35.5 cm2      (Approx.)Area(ABCD)    =ar(ΔBCD)+ar(ΔABD)    =30+35.5    =65.5 cm2Thus, area of ABCD is 65.5 cm2 approx.

Q.2 Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Ans

In ΔABC,3,4 and 5 are Pythagorian triplet.So, area(ΔABC)=12×BC×AB                     =12×4×3                     =6 cm2Area of ΔACD,by Pythagoras Theorem,                  Δ=s(s−a)(s−b)(s−c)                   s=5+5+42                      =142=7 cm                  Δ=7(7−5)(7−5)(7−4)                    =7(2)(2)(3)                     =9.2 (Approx.)∴Area(ABCD)=area(ΔABC)+area(ΔACD)                   =6+9.5                   =15.2 cm2      (Approx)

Q.3 Radha made a picture of an aeroplane with coloured paper as shown in the following figure. Find the total area of the paper used.

Ans

For I triangle:sides are 5 cm, 1 cm and 5 cm.By Heron’s formula:Area of triangle(Δ)=s(s−a)(s−b)(s−c)where,​              s=a+b+c2                       =5+1+52=5.5 cm∴                     Δ=5.5(5.5−5)(5.5−1)(5.5−5)                       =5.5(0.5)(4.5)(0.5)                        =2.488 cm2Area of II figure:   Area of rectangle=Length×Breadth                        =6.5×1                        =6.5 cm2Area of III figure:

Height of ΔBCF=12−(0.5)2                    =1−0.25                    =0.75                    =0.866  cmArea of trapezium ABCD                    =12(sum of parallel sides)×height                    =12(1+2)×0.866                    =1.299 cm2Area of IV figure:Area of right triangle                    =12×1.5×6                    =4.5Area of V figure:Area of right triangle                    =12×1.5×6                    =4.5Total area of used paper=2.488+6.5+1.299+4.5+4.5                    =19.287 cm2                    =19.3 cm2 (Approx.)

Q.4 A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Ans

Sides of triangle are 26 cm, 28 cm and 30 cm.Area of triangle by Heron’s fromula,                Δ=s(s−a)(s−b)(s−c)                 s=26+28+302=42                Δ=42(42−26)(42−28)(42−30)                   =42(16)(14)(12)                   =336 cm2Let height of parallelogram be h cm.Since, area of parallelogram=Area of triangle                           base×h=336 cm2                             28×h=336 cm2                                  h=33628=12 cmThus, height of parallelogram is 12 cm.

Q.5 A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Ans

Sides of triangle are 30 m, 30 m and 48 m. By Heron’s formula: Area ( ABC )= s( s−a )( s−b )( s−c )              s= a+b+c 2                = 30+30+48 2 =54 Area ( ABC )= 54( 54−30 )( 54−30 )( 54−48 )                = 54( 24 )( 24 )( 6 )                =432 cm 2 Area of rhombus ABCD                =2×Area( ΔABC )                =2×432                =864  m 2             Number of cows=18 Area grazed by each cow= 864 18                                =48  m 2 Thus, the area of grass grazed by each cow is 48  m 2 of the field. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@26D9@

Q.6 An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Ans

         Number of triangular pieces in umbrella=10Number of triangular pieces of each coulour=5       Sides of each triangle=20 cm, 50 cm, 50 cmarea of triangle, by Heron’s fromula                             Δ=s(s−a)(s−b)(s−c)where,         s=a+b+c2=20+50+502=60                             Δ=60(60−20)(60−50)(60−50)                             Δ=60(40)(10)(10)=2006 cm2Area of required cloth for each colour in umbrella                               =5×2006                                =10006  cm2Thus, the cloth required for one colour is 10006 cm2 and clothrequired for other colour is 10006 cm2.

Q.7 A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Figure. How much paper of each shade has been used in it?

Ans

Area of square =12×Product of diagonals                    =12×32×32                    =512 cm2Since, paper required for area of shade I                     =paper required for area of shade II                    =5122cm2                    =216 cm2Thus, the required paper of each part I and II is 216 cm2.Sides of triangle are 6cm, 6 cm and 8 cm.Area of triangle=s(s−a)(s−b)(s−c)       [By Heron’s formula]                    =10(10−6)(10−6)(10−8)                                    [∵s=6+6+82=10]Area of triangle=10(4)(4)(2)                    =85                    =8×2.24Area of triangle=17.92 cm2Thus, the area of required paper for I and II each part is 216 cm2 and that of for III shaded part is 17.92 cm2.

Q.8 A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm.
Find the cost of polishing the tiles at the rate of 50p per cm2.

Ans

Sides of triangle used in floral design are 9 cm, 28 cm and 35 cm.So, area of a triangle=s(s−a)(s−b)(s−c)                           =36(36−9)(36−28)(36−35)                                   [∵s=9+28+352 =36 cm]                           =36(27)(8)(1)                           =366 cm2                           =36×2.45cm2                           =88.2 cm2    Area of 16 triangles=16×88.2                           =1411.2 cm2Cost of polishing1 cm2 area of floor                           =₹ 50100                           =₹0.5Cost of polishing 1411.2 cm2 area of floor                           =₹0.5×1411.2                           =₹705.6Thus, the cost of polishing the tiles at the rate of 50p per cm2 is₹705.6.

Q.9 A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Ans

Let ABCD be trapezium and parallel sides AB=10 m and CD = 25 m.
Non-parallel sides are AD = 13 m and BC = 14 m.
Let height of ABCD be BE. Draw BF parallel to AD.
So, BF = 13 cm because opposite sides of parallelogram are equal.
FC = 25 – DF
= 25 – 10 = 15 m

Sides of ΔBFC are 13 m, 15 m and 14 m.Area of ΔBFC=s(s−a)(s−b)(s−c)          [By Heron’s formula]                  =21(21−13)(21−15)(21−14)                                 [∵s=13+15+142=21 m]                 =21(8)(6)(7)                  =7056                 =84​ m2Area of ΔBFC=12×FC×BE                 84​ m2=12×15×BE⇒height,  BE=2×8415=11.2 m∴area(ABCD)=12(AB+CD)×BE         [∵area of trapezium=12(sum of parallel sides)×h]area(ABCD)=12(10+25)×11.2                 =12(35)×11.2                =196 m2

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