# NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.2) Exercise 11.2

The area calculation for Parallelograms and Triangles is covered in detail in the NCERT solutions for Mathematics Chapter 11 Perimeter and Area Exercise 11.2 for Class 7 students. It has also been rationally demonstrated here how triangles are a component of a rectangle. Students are required to compute the area and perimeter of Triangles and Parallelograms as well as, in certain circumstances, locate the missing sides, as part of the practise problems. There are 8 questions in total, the solutions to all of which have a straightforward methodology, in the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 based on Perimeter and Area. The purpose of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2  is to give students a step-by-step method for solving difficult mathematical problems and to help them locate solutions to their problems. Students may review the complete curriculum and raise their grades by using the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2. Students who are studying the area of Triangles will notice that if two Triangles exist and are congruent, their areas will be equal, but if their areas are equal, two Triangles may not be congruent.

The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 require students to demonstrate their understanding of Parallelograms and types of Triangles. To practice Class 7 Maths Ch 11 Ex 11.2,Before attempting to calculate the area and perimeter of any geometric figure in Class 7 Maths Ch. 11 Ex. 11.2, it is necessary to have a thorough understanding of the fundamental concepts of all geometric figures. Due to the fact that all the questions are based on formulae, students must be familiar with all the formulas given in the chapter for  NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 area and perimeter of parallelograms and different types of triangles. There are several exercises in this chapter that contain multiple questions in addition to the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2. As discussed previously, the subject matter specialists have already resolved or responded to all of these questions. Due to this, they are all guaranteed to be of the highest quality, and anybody may use them to study for exams. It is essential to understand all the ideas in the textbooks and work through the exercises that are provided to receive excellent results in the annual exams.

## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.2) Exercise 11.2

For CBSE students preparing for exams, using the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2  is seen to be the best alternative. There are several exercises in this chapter. On this page, in PDF format, Extramarks offers the Class 7 Maths Chapter 11 Exercise 11.2 Solution. Students can study the NCERT Solutions For Class 7 Maths Chapter 11 Exercise 11.2 straight from the Extramarks website or mobile application, or they can download it as needed. Extramarks has engineered NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 PDF Download. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 are holistic reference materials for understanding all of the chapter’s subjects.

### Access NCERT Solutions for Class 7 Chapter 11 – Perimeter and Area

The study of all the shapes that surround humans in detail is covered by the mathematical subdiscipline of Geometry; this study mostly entails calculating the areas and perimeters of those shapes. Due to their importance in humans’ daily lives, these parameters are crucial to compute. The ideas of area and perimeter of various shapes, including Square, Rectangle, Parallelogram, Triangle, and Circle, are covered in detail in the NCERT solutions for class 7 Mathematics Chapter 11 Perimeter and Area. When it comes to the area and perimeter of forms, knowing units is equally essential. For this reason, the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 Perimeter and Area also explains how to convert lengths and areas into other units. To assist with computations, students must understand that although area refers to the space around a form, perimeter refers to the length of that shape’s perimeter. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 are designed to assist students to achieve the highest possible grade in their exams. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 feature several important pictures to assist students in better understanding the material. Students could use the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 to fully comprehend all of the topics in the textbooks and finish the questions. These study resources might help students in achieving excellent scores on exams. Students have learned the perimeters of plane figures and the areas of Squares and Rectangles in Class 6. A closed figure perimeter is the space around it, but its area is the portion of a plane or region that it takes up. Students will study the areas and perimeters of a few more plane figures in this session. The NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.2 are reliable exam preparation resources.It is recommended that students read through these NCERT solutions for Class 7. The solutions are all offered step-by-step and in compliance with the CBSE guidelines so that students may practise while understanding the reasoning behind each problem.

Chapter 1 Integers

The study of integers is covered in Chapter 1. Whole numbers and integers are concepts that students learned about in earlier lessons. Students will now learn more about numbers, their characteristics, and their uses. Similar concepts apply to the addition and subtraction of numbers, the characteristics of addition and subtraction of integers, and the multiplication and division of integers.

Chapter 2 Fractions and Decimals

The NCERT textbook Chapter 2 covers fractions and decimals. In prior sessions, students learned about fractions and decimals as well as how to add and subtract from them. Students are now learning how to multiply and divide decimals and fractions.

Chapter 3 Data Handling

In Chapter 3 of the NCERT textbook, data handling is covered. Students have worked with a variety of data types in prior lessons. Students now know how to gather data, tabulate it, and present it as bar graphs. Students will now take a further step toward understanding how to achieve this in this chapter. Data collection, data organisation, representative values, arithmetic mean, mode of huge data, median, usage of bar graphs for various purposes, scale selection, chance, and probability are all covered in this chapter.

Chapter 4 Simple Equations

Setting up an equation and solving an equation are covered in Chapter 4 of the NCERT textbook. Students will learn how to transpose a number or move it from one side to the other, as they go through more equations. Instead of adding or subtracting integers from both sides of an equation, students can transpose the numbers.

Chapter 5 Lines and Angles

Lines and angles are the topics of this chapter. This chapter discusses transversals, the angle created by a transversal, the transversal of parallel lines, and the checking of parallel lines. It also discusses related angles, complementary angles, supplementary angles, adjacent angles, linear pairs, vertically opposite angles, pairs of lines intersecting, and transversals.

Chapter 6 The Triangle and its Properties

The Triangle and Its Properties is the topic of Chapter 6 of the NCERT textbook. A simple closed curve made up of three line segments is a triangle. The object has three sides, three angles, and three vertices. The notions of a Triangle’s median, altitude, exterior angle and its property, angle sum property, two special Triangles—equilateral and isosceles—the sum of its two sides, right-angled triangles, and Pythagoras property—are all covered.

Chapter 7 Congruence of Triangles

The congruence of triangles is the subject of this chapter. Congruent refers to two figures that are precisely the same size and shape. Congruence of planar figures, congruence of line segments, congruence of angles, congruence of triangles, requirements for congruence of triangles, and congruence of right-angled triangles are more subjects connected to this one.

Chapter 8 Comparing Quantities

In the NCERT textbook, Chapter 8, the topic of comparing quantities is covered. Equivalent ratios, percentage as a different approach to comparing quantities, the definition of percentage, converting fractional values to percentages, translating decimal numbers to percentages, and converting percentages to fractions or decimals are some of the subjects that are linked to it. Other fascinating themes include estimating games, percentage interpretation, ratio conversion, increase or decrease expressed in percentages, profit or loss expressed in percentages, charges applied to borrowed funds, and simple interest.

Chapter 9 Rational Numbers

The topic of rational numbers is covered in Chapter 9 of the NCERT textbook. Besides covering positive and negative rational numbers, rational numbers on a number line, rational numbers in standard form, rational number comparisons, rational number comparisons between two rational numbers, and operations on rational numbers, it also covers these topics in greater detail.

Chapter 10 Practical Geometry

The theme of Practical Geometry is covered in Chapter 10 of the NCERT textbook. Students are familiar with several forms. Some of these are drawn by them during prior lessons. They can draw a line segment of a certain length, a line perpendicular to a certain line segment, an angle, an angle bisector, a circle, etc. as examples. They will now learn how to draw various kinds of triangles and parallel lines. Highly qualified mentors have compiled the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 to cover all of the major themes incorporated in Chapter 11.

Chapter 11 Perimeter and Area

The chapter introduces the Mensuration portion of the curriculum. It deals with the perimeters and areas of all the significant mathematical forms. The chapter is pretty straightforward and doesn’t introduce any complicated forms. To properly understand the material, students are advised to refer to the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2

Chapter 12 Algebraic Expressions

Finding the values of an expression is covered in Chapter 12 of the NCERT textbook. This chapter also covers terms of an expression, coefficients, like and unlike terms, monomials, binomials, trinomials, and polynomials. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 are presented in a simple, clear, and concise manner.

Chapter 13 Exponents and Powers

The topic of Exponents and Powers is covered in Chapter 13 of the NCERT textbook. Exponents are the result of multiplying rational integers several times on their own. The rules of exponents, multiplying and dividing powers with the same base, taking the power of a power, multiplying and dividing powers with the same exponents, the decimal number system, and expressing huge numbers in standard form are some of the topics discussed in this chapter.

Chapter 14 Symmetry

Symmetry is the topic of this chapter. When a form is moved in some way—for example, by turning, flipping, or sliding—it becomes precisely like another shape. Two items must be identical in terms of size, form, and orientation for them to be symmetrical concerning one another. A face is an example of an item that can have symmetry. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 provide answers to all of the practice problems that are included at the end of Chapter 11. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 aid students in quickly and viably grasping the large wide range of concepts.

Chapter 15 Visualising Solid Shapes

Drawing solids on a flat surface, oblique drawings, isometric sketches, faces, edges, and vertices are all covered in Chapter 15 of the NCERT textbook, “Visualising Solid Shapes.”

### NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2

In accordance with the criteria in the NCERT books, qualified educators have provided the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 and solutions to all chapters’ exercises cumulatively on the Extramarks website. The NCERT Class 7 Maths Chapter 11 Exercise 11.2 will help students  have a firm grasp of the concepts in the chapter. For students studying for academic exams, the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 are regarded as the best choice. Exercise 11.2 Class 7th contains lots of questions. Students can face difficulties while completing Class 7 Math Ch 11 Ex 11.2. Students can get the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 on the Extramarks website. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 are logically sequenced and error-free. Students can get the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 directly from the Extramarks website or mobile application, or they can download them if necessary. They can also learn an alternative, easier way to solve a particular geometrical problem. All of the questions cannot be answered in one class. The answers to all of the sums in NCERT Class 7 Maths Chapter 11 Exercise 11.2 can be found in the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2.

Students can enhance their theoretical and practical understanding by studying the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 are available as a PDF download and cover every question from Class 7th Math Exercise 11.2. There are suggestions and pointers in addition to the solutions. That would make it easier to respond to queries. On Extramarks, students can access the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 in PDF format.

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Q.1 Find the area of each of the following parallelogram :

Ans

$\begin{array}{l}\text{(a)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{4 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{7 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{7}×\text{4}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{28cm}}^{2}\\ \text{(b)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{3 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}×\text{3}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{15 cm}}^{2}\\ \text{(c)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{3}\text{.5 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{2}\text{.5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}\text{.5}×\text{3}\text{.5}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8}{\text{.75 cm}}^{2}\end{array}$ $\begin{array}{l}\text{(d)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{4}\text{.8 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}×\text{4}\text{.8}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{24 cm}}^{2}\\ \text{(e)}\\ \text{Area of Parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height}=\text{4}\text{.4 cm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Base}=\text{2 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}×\text{4}\text{.4}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8}{\text{.8 cm}}^{2}\end{array}$

Q.2 Find the area of each of the following triangles :

Ans

$\begin{array}{l}\text{(a) Area of a triangle}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×4×3\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×12=6{\text{cm}}^{2}\\ \text{(b) Area of a triangle}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×5×3.2\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×16=8{\text{cm}}^{2}\\ \text{(c) Area of a triangle}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×3×4\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×12=6{\text{cm}}^{2}\\ \text{(d) Area of a triangle}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×3×2\\ \text{}\text{}\text{}\text{}=\frac{1}{2}×6=3{\text{cm}}^{2}\end{array}$

Q.3 Find the missing values :

 S. NO. Base Height Area of the Parallelogram a. 20 cm 246 cm2 b. 15 cm 154.5 cm2 c. 8.4 cm 78.72 cm2 d. 15.6 cm 16.38 cm2

Ans

$\begin{array}{l}\text{Let the height be}h\text{and base be}b.\\ \text{(a) Area of parallelogram=Base}×\text{Height}\\ \text{So,we get}\\ {\text{246 cm}}^{\text{2}}=\text{20 cm}×\text{h}\\ \text{h=}\frac{246\text{\hspace{0.17em}}{\text{cm}}^{2}}{20\text{\hspace{0.17em}}\text{cm}}\\ =12.3\text{\hspace{0.17em}}\text{cm}\\ \text{(b) Area of parallelogram=Base}×\text{Height}\\ \text{So,we get}\\ \text{154}{\text{.5 cm}}^{\text{2}}=\text{b}×\text{15 cm}\\ \text{b=}\frac{154.5\text{\hspace{0.17em}}{\text{cm}}^{2}}{15\text{\hspace{0.17em}}\text{cm}}\\ =10.3\text{\hspace{0.17em}}\text{cm}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\text{(c) Area of parallelogram=Base}×\text{Height}\\ \text{So,we get}\\ \text{78}{\text{.72 cm}}^{\text{2}}=\text{b}×\text{8}\text{.4 cm}\\ \text{b=}\frac{48.72\text{\hspace{0.17em}}{\text{cm}}^{2}}{8.4\text{\hspace{0.17em}}\text{cm}}\\ =5.8\text{\hspace{0.17em}}\text{cm}\\ \text{(d) Area of parallelogram=Base}×\text{Height}\\ \text{So,we get}\\ \text{16}{\text{.38 cm}}^{\text{2}}=\text{15}\text{.6 cm}×\text{h}\\ \text{h=}\frac{16.38\text{\hspace{0.17em}}{\text{cm}}^{2}}{15.6\text{\hspace{0.17em}}\text{cm}}\\ =1.05\text{\hspace{0.17em}}\text{cm}\\ \text{So, we get}\\ \begin{array}{cccc}\text{S}\text{.No}\text{.}& \text{Base}& \text{Height}& \text{Area of the Parallelogram}\\ \text{a}\text{.}& 20\text{cm}& 12.3\text{\hspace{0.17em}}\text{cm}& 246{\text{cm}}^{2}\\ \text{b}\text{.}& 10.3\text{\hspace{0.17em}}\text{cm}& 15\text{cm}& 154.5{\text{cm}}^{2}\\ \text{c}\text{.}& 5.8\text{\hspace{0.17em}}\text{cm}& 8.4\text{cm}& 78.72{\text{cm}}^{2}\\ \text{d}\text{.}& 15.6\text{cm}& 1.05\text{\hspace{0.17em}}\text{cm}& 16.38{\text{cm}}^{2}\end{array}\end{array}$

Q.4 Find the missing values :

 Base Height Area of Triangle 15 cm …………….. 84 cm2 …………….. 31.4 mm 1256 mm2 22 cm …………….. 170.5 cm2

Ans

$\begin{array}{l}\text{(a) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}×\text{base}×\text{height}\\ \text{}\text{}\text{\hspace{0.17em}}{\text{87cm}}^{2}=\frac{1}{2}×15\text{\hspace{0.17em}}\text{cm}×\text{h}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{h}=\frac{87{\text{cm}}^{2}×2}{15\text{\hspace{0.17em}}\text{cm}}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=11.6\text{cm}\\ \text{(b) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}×\text{base}×\text{height}\\ \text{}\text{}\text{\hspace{0.17em}}{\text{1256 mm}}^{2}=\frac{1}{2}×b×\text{31}\text{.4 mm}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=\frac{1256{\text{mm}}^{2}×2}{31.4\text{\hspace{0.17em}}\text{mm}}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=80\text{mm}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\text{(c) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}×\text{base}×\text{height}\\ \text{}\text{}\text{\hspace{0.17em}}\text{170}{\text{.5 cm}}^{2}=\frac{1}{2}×22\text{\hspace{0.17em}}\text{cm}×\text{h}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{h}=\frac{170.5{\text{cm}}^{2}×2}{22\text{\hspace{0.17em}}\text{cm}}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=15.5\text{cm}\\ \text{So we get}\\ \begin{array}{ccc}\text{Base}& \text{Height}& \text{Area of Triangle}\\ 15\text{cm}& 11.6\text{cm}& 84{\text{cm}}^{2}\\ 80\text{\hspace{0.17em}}\text{mm}& 31.4\text{mm}& 1256\text{\hspace{0.17em}}{\text{mm}}^{2}\\ 22\text{cm}& 15.5\text{\hspace{0.17em}}\text{cm}& 170.5{\text{cm}}^{2}\end{array}\end{array}$

Q.5

Ans

$\begin{array}{l}\text{(a)}\\ \text{Area of a parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{SR}×\text{QM}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=7.6×12\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=91.2\text{\hspace{0.17em}}{\text{cm}}^{2}\\ \text{(b)}\\ \text{Area of a parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{PS}×\text{QN}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=91.2\text{\hspace{0.17em}}{\text{cm}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{QN}×8=91.2\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{QN}=\frac{91.2}{8}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=11.4\text{\hspace{0.17em}}\text{cm}\end{array}$

Q.6

$\begin{array}{l}\mathrm{DL}\mathrm{and}\mathrm{BM}\mathrm{are}\mathrm{the}\mathrm{heights}\mathrm{on}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{AD}\mathrm{respectively}\\ \mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}.\mathrm{If}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}\mathrm{is}\\ 1470{\mathrm{cm}}^{\mathrm{2}},\mathrm{AB}= 35\mathrm{cm}\mathrm{and}\mathrm{AD}= 49\mathrm{cm},\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\\ \mathrm{BMand}\mathrm{DL}.\end{array}$

Ans

$\begin{array}{l}\text{Area of parallelogram}=\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{AB}×\text{DL}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1470}=\text{35}×\text{DL}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{DL}=\frac{1470}{35}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=42\text{\hspace{0.17em}}\text{cm}\\ \text{Also,}\\ 1470=\text{AD}×\text{BM}\\ \text{1470}=\text{49}×\text{BM}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{BM}=\frac{1470}{49}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=30\text{\hspace{0.17em}}\text{cm}\end{array}$

Q.7

$\begin{array}{l}△\mathrm{ABC}\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{A}.\mathrm{AD}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{BC}.\\ \mathrm{If}\mathrm{AB}=5\mathrm{cm}, \mathrm{BC}=13\mathrm{cm}\mathrm{and}\mathrm{AC}=12\mathrm{cm},\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\\ △\mathrm{ABC}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{AD}.\end{array}$

Ans

$\begin{array}{l}\text{Since, Area}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}×\text{5}×\text{12}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{30 cm}}^{2}\\ \text{Also, area of triangle}=\frac{1}{2}×\text{AD}×\text{BC}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{30}=\frac{1}{2}×\text{AD}×\text{13}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{AD}=\frac{30×2}{12}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.6\text{\hspace{0.17em}}\text{cm}\end{array}$

Q.8

$\begin{array}{l}△\mathrm{ABC}\mathrm{is}\mathrm{isosceles}\mathrm{with}\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}\mathrm{and}\mathrm{BC}=9\mathrm{cm}.\\ \mathrm{The}\mathrm{height}\mathrm{AD}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{BC},\mathrm{is}6\mathrm{cm}.\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}△\mathrm{ABC}.\\ \mathrm{What}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{C}\mathrm{to}\mathrm{AB}\mathrm{i}.\mathrm{e}.,\mathrm{CE}?\end{array}$

Ans

$\begin{array}{l}\text{Area of}\Delta \text{ABC}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}=\frac{1}{2}\text{BC}×\text{AD}=\frac{1}{2}×\text{9}×\text{6}={\text{27cm}}^{\text{2}}\\ \text{Also, area of}\Delta \text{ABC}=\frac{1}{2}×\text{Base}×\text{Height}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\text{AB}×\text{CE}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}27=\frac{1}{2}\text{7}\text{.5}×\text{CE}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{CE}=\frac{27×2}{7.5}=\text{7}\text{.2 cm}\end{array}$

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### 1. How many questions are there in Class 7th Exercise 11.2?

There are 8 questions in Maths Class 7 Chapter 11 Exercise 11.2. Students who use the methods depicted in NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 would perform well in their exams.

### 2. Are the Class 8 Mathematics Chapter 2 NCERT solutions in Hindi sufficient for test preparation?

A great resource for Class 8 Mathematics studies is the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2. Numerous textbook allusions do not imply that students receive in-depth learning or any other advantages. The fact that the solutions are presented differently in each book, can raise even more scepticism. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2 are the best study aid that students can use to prepare for examinations. Students will not want anything more if they are provided with the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2.

### 3. What are the important formulas of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2?

There are lots of crucial formulas present in the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.2. These formulas will help while solving the Class 7 Math Chapter 11 Exercise 11.2

The perimeter of a square = 4 × side

Perimeter of a rectangle = 2 × (length + breadth)

Area of a square = side × side

Area of a rectangle = length × breadth

Area of a parallelogram = base × height

Area of a triangle = ½ × base × height

Circumference of a circle = 2πr

Area of a circle = πr2