# NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.3) Exercise 11.3

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**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.3) Exercise 11.3**

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**Access Ncert Solution for Class 7 Chapter 11 – Perimeter and Area **

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The Class 7th Maths Exercise 11.3 is one of the sub topics of Geometry. Geometry is a discipline of Mathematics that largely deals with the shapes and sizes of objects, their relative positions, and the qualities of space (from the Ancient Greek: geo- “earth,” -metron “measuring”). The Greek mathematician Euclid, commonly referred to as the “Father of Geometry,” used numerous postulates and theorems. Students are advised to use tools such as the NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.3 for Class 7 Math Chapter 11.3 to better understand the geometry subtopics.

The domain of Mathematics known as Geometry connects the concepts relating to areas, Volumes, Patterns, Angles, and Distances. Geometry is the umbrella term for all concepts that are visually and spatially related. There are three different types of geometry:

- Euclidean
- Hyperbolic
- Elliptical

Elliptical Geometry –

To comprehend the basics of Geometry, students study Euclidean geometry. Axioms and theorems are the foundation of Euclidean geometry, which is the study of solid and plane figures. Points and Lines, Euclid’s Axioms and Postulates, the Geometrical Proof, and Euclid’s Fifth Postulate are some of the basic ideas of Euclidean geometry. The five Euclidean Geometry postulates are used to define geometrical figures.

- Any point to any other point is connected by a section of a straight line.
- The length of a straight line is infinite in both directions.
- Any specified location can serve as the circle’s centre and any length can serve as the radius.
- Congruent angles are all just angles.
- Any two straight lines that are equally distant from one another at two places are eternally parallel.

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The two non-Euclidean geometries are Spherical Geometry and Hyperbolic Geometry. In contrast to Euclidean Geometry, non-Euclidean Geometry has different hypotheses about the Nature of Parallel Lines and Angles in flat space.

- The study of Sphere-Based Plane Geometry is known as Spherical Geometry. The shortest distance between two places along a line is how it is defined. The great circle is the name given to this Arc-Shaped Line on a Sphere. The triangle’s total angles are larger than 180 degrees.
- A Curved Surface is referred to as Hyperbolic Geometry. Topology is where this Geometry is used. The planar triangle has a total of the angles that is less than 180 degrees, depending on the interior curvature of the curved surface.

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Plane Geometry –

The study of Geometry in a plane is a component of Euclidean Geometry. The plane is a two-dimensional surface that can go on forever in both directions. Every aspect of Geometry and graph theory uses Planes. Points, Lines, and Angles are equivalent to the fundamental elements of Planes in Geometry. The fundamental geometric unit with no dimensions is the point. The Collinear Points are those that are on the same line. A line is a single-dimensional object that denotes a collection of points extending in two opposing directions; it is defined as the place where two planes join. There are no ends to a Line. A Line, Line Segment, and ray may all be distinguished easily. Lines can be Perpendicular or Parallel. Lines may or may not cross.

This is the type of Geometry that is taught initially to lower classes and Extramarks has tools that are created for those particular classes as well, along with the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3.

Solid Geometry-

Geometric solids have a three-dimensional character. Length, width, and height are the three dimensions that are taken into account. Solid shapes come in many varieties, including Cylinders, Cubes, Spheres, Cones, Cuboids, Prisms, Pyramids, and so on. These shapes take up space. They have vertices, faces, and edges that define them. In Euclidean space, the Polyhedrons and the five platonic solids have intriguing characteristics.

This is the type that’s very common for middle school students. They can practise this theme with the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other similar tools.

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The greatest method to comprehend any chapter of Mathematics is to practise as many problems as possible. Students can easily complete these solutions by going through the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. Students’ chances of mastering subjects that are typically termed as complex and difficult are increased as they practise using resources like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 are advised to be thoroughly read through and used frequently while practising for exams. Use of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 should not be delayed until after exams. These learning tools should be used for both homework and other assignments, such as tests in class.

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The Math Class 7th Exercise 11.3 is about Perimeter and area. The whole length of any closed shape boundary is known as its Perimeter. Students might have a sizable square-shaped farm, for instance. Students now decide to fence their farm to protect it from stray animals. Finding the total length of the farm’s boundary is as simple as multiplying the length of one side of the farm by 4. There are a lot of situations like this when students can be applying the Perimeter-Finding Notion without even realising it. The use of NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 can help students, better understand this concept. These are a few basics that are part of the topic taught under Chapter 11 of Class 7 Mathematics. These can all be better understood with the help of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 .

The whole distance encircling a shape is referred to as its Perimeter. It is the length of any two-dimensional geometric shapes boundary or outline. Depending on the size, the perimeter of several figures can be the same. Consider a triangle built of an L-length wire, for instance. If all the sides are the same length, the same wire can be used to create a square. Students are advised to make use of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 so that they do not get overwhelmed by this sub-topic of Geometry and get good practise with it.

**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3 **

All vocations in the world depend on Mathematics. The application of Mathematics and numbers to many facets of life is crucial. It is used to advance and explain the theories of other branches of Science, particularly those in Physics, Chemistry, Astronomy, Geography, etc. It allows intellectuals to conduct several experiments to test their theories. Extramarks provides tools like the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3. The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other similar tools can help students master important subjects like Mathematics. Tools such as the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 available on the Extramarks website can help students master Mathematics and make a career in various fields.

Everyone owns a cell phone, which requires a basic understanding of abilities and numbers. One must be familiar with the functions of the phone’s numbers. With today’s technology, using a cell phone for anything from talking and chatting to Internet browsing is possible.

Mathematics also aids in the learning of data entry in many fields that only depend on Mathematics, including Engineering, Physics, Computer Programming, Accounting, and Banking. An essential tool for computing is Mathematics.

The core of computer operation is binary Mathematics. The most fundamental form of Mathematics utilised in Computer Science to represent each number in the machine is a binary number. Students can make the most of the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 and other tools available on the Extramarks website to better understand these concepts that have their basis in Mathematics. Algebra, Statistics, and Calculus I, and Calculus II are some of the mathematical procedures frequently utilised in computer programming. The use of tools created by the experts at Extramarks that are targeted towards Mathematics, for example the NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 can help students understand the real life uses of Mathematics

Technical innovation: Mathematics is crucial to the creation of contemporary technological tools like the Internet, which is a necessity of daily life and makes it easier for people to communicate, find information, and navigate.

Mathematics is frequently employed in animation design. When working with moving and changing objects, this enables the animator to extrapolate features of geometric figures and discover the unknown from a straightforward set of equations. The animation designer demonstrates how the designs are animated, transformed, and zoomed in and out using Linear Algebra. Students interested in these fields should have a strong foundation in mathematics, which they can achieve by using the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and other similar tools available on the Extramarks website.

Mathematics is a key element in all engineering disciplines and is extensively employed in Architecture. Mathematics is used by architects to determine the square footage of rooms and structures as well as the measurements of the ground area and the space needed for additional areas like parking, plumbing, and other areas. A lot of students go into this field after completing their schooling. Students are advised to develop a healthy relationship with Mathematics in order to have a successful career in this field. They can do so by using resources such as the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and many other tools available on the Extramarks website.

An essential component of the sports world is Mathematics. It significantly affects how effectively sports are played. A person’s cognitive and decision-making abilities are enhanced by Mathematics. These abilities are crucial for a sportsperson because they enable him to choose the best course of action for his squad. The player can choose the path and angle the ball will take to reach the goal with the aid of Engineering and Trigonometry. A person who struggles with Mathematics may find it difficult to estimate situations accurately and make wise choices.

This is a field that many students find very closely related to Mathematics. Students should try to look at such real-life implementations of Mathematics in order to develop a liking for the subject and not fear it. They can improve their performance in the subject by using the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3, and other tools created by Extramarks’ mathematics experts for students.

When painting, Spatial Calculations are crucial. Calculating how much paint is required to complete a task, such as painting a wall, is necessary. In 3D art, mathematical calculations are used to decide how to create flat objects so that they appear three-dimensional from a particular vantage point. When a particular angle is taken when looking at 2D flat art, the proportions are warped to create the appearance of a 3D item. This may come as a shock to students. This is how important Mathematics is. It is even a part of fields like Art and Painting which many might conventionally consider the opposite of Mathematics.

There are even more parts of real life that are closely associated with mathematics. Understanding mathematics is always an added skill that helps in leading a better life. Students are advised to try to understand the subject at a young age.Most students get overwhelmed by the subject because of their lack of background knowledge. This is the reason why many students dislike or fear the subject of Mathematics.

Such problems can and should be handled at an early age. The students can do so with the help of the expert-made resources available on the Extramarks website. These are available for elementary classes as well. The NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3 is just one of the many educational resources for Mathematics available on the Extramarks website.

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The NCERT Solutions Class 7 Maths Chapter 11 Exercise 11.3 has step-by-step solutions to the questions that are asked in the 11.3 exercise of Chapter 11 in the Class 7 Mathematics book. Students can find extra questions and their solutions on the Extramarks website easily. However, the NCERT Solutions for Class 7 Maths, Chapter 11, Exercise 11.3 only address the questions found in NCERT textbooks.

**Q.1 **

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{circumference}\mathrm{of}\mathrm{the}\mathrm{circles}\mathrm{with}\mathrm{the}\mathrm{following}\\ \mathrm{radius}:\left(\mathrm{Take}\mathrm{\pi}\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{a}\right)14\mathrm{cm}\left(\mathrm{b}\right)28\mathrm{mm}\left(\mathrm{c}\right)21\mathrm{cm}\end{array}$

**Ans**

\begin{array}{l}\text{(a)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\times \frac{22}{7}\times 14\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\times 22\times 2\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=88\text{\hspace{0.17em}}\text{cm}\\ \text{(b)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\times \frac{22}{7}\times 28\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\times 22\times 4\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=176\text{\hspace{0.17em}}\text{mm}\end{array}

\begin{array}{l}\text{(c)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\times \frac{22}{7}\times 21\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\times 22\times 3\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=132\text{\hspace{0.17em}}\text{cm}\end{array}

**Q.2 **

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{circles},\mathrm{given}\mathrm{that}:\\ \left(\mathrm{a}\right)\mathrm{radius}=\; 14\mathrm{mm}\left(\mathrm{Take}\mathrm{\pi}\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{b}\right)\mathrm{diameter}=\; 49\mathrm{m}\left(\mathrm{c}\right)\mathrm{radius}=\; 5\mathrm{cm}\end{array}$

**Ans**

\begin{array}{l}\text{(a)}\\ \text{Area of a circle}=\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}{\left(14\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=616\text{\hspace{0.17em}}{\text{mm}}^{2}\end{array}

\begin{array}{l}\text{(b)}\\ \text{Radius}=\frac{\text{Diameter}}{2}=\frac{49}{2}=24.5\text{\hspace{0.17em}}\text{mm}\\ \text{Area of a circle}=\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}{\left(24.5\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1886.5\text{\hspace{0.17em}}{\text{m}}^{2}\\ \text{(c)}\\ \text{Area of a circle}=\pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}{\left(5\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{550}{7}\text{\hspace{0.17em}}{\text{cm}}^{2}\end{array}

**Q.3 ****If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =22/7)**

**Ans**

\begin{array}{l}\text{Since, circumference}=\text{2}\pi \text{r}\\ \text{So, we get}\end{array}

\begin{array}{l}\text{154}=\text{2}\times \frac{22}{7}\times r\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=\frac{154\times 7}{2\times 22}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{49}{2}\text{\hspace{0.17em}}\text{m}=4.5\text{\hspace{0.17em}}\text{m}\\ \text{Now,}\\ \text{Area}=\pi {\text{r}}^{2}\\ \text{}\text{\hspace{0.17em}}=\frac{22}{7}\times \frac{49}{2}\times \frac{49}{2}\\ \text{}\text{\hspace{0.17em}}=1886.5\text{\hspace{0.17em}}{\text{m}}^{2}\end{array}

**Q.4 **

$\begin{array}{l}\mathrm{A}\mathrm{gardener}\mathrm{wants}\mathrm{to}\mathrm{fence}\mathrm{a}\mathrm{circular}\mathrm{garden}\mathrm{of}\mathrm{diameter}\\ 21\mathrm{m}.\mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rope}\mathrm{he}\mathrm{needs}\mathrm{to}\mathrm{purchase},\mathrm{if}\mathrm{he}\\ \mathrm{makes}2\mathrm{rounds}\mathrm{of}\mathrm{fence}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{costs}\mathrm{of}\mathrm{the}\mathrm{rope},\mathrm{if}\mathrm{it}\\ \mathrm{cost}\u20b9\; 4\mathrm{per}\mathrm{meter}.\left(\mathrm{Take\; \pi}\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}$

**Ans**

\begin{array}{l}\text{Given,}d=\text{21 m}\\ \text{So,}r=\frac{21}{2}\text{\hspace{0.17em}}\text{m}\\ \text{Circumference}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\times \frac{22}{7}\times \frac{21}{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=66\text{\hspace{0.17em}}\text{m}\\ \text{Length of the rope required for fencing}=\text{2}\times \text{66 m}\\ \text{}\text{}\text{}\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{132 m}\\ \text{}\text{\hspace{0.17em}}\text{Cost of 1 m rope}=\text{\u20b94}\\ \text{So, cost of 132 m rope}=\text{\u20b9 4}\times \text{132}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{\u20b9528}\end{array}

**Q.5 ****From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)**

**Ans**

\begin{array}{l}\text{Outer radius of circular sheet}=\text{4 cm}\\ \text{Inner radius of circular sheet}=\text{3 cm}\end{array}

\begin{array}{l}\text{Remaining area}=-\left(3.14\times 3\times 3\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=50.24-28.26\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=21.98\text{\hspace{0.17em}}{\text{cm}}^{2}\end{array}

**Q.6 **

$\begin{array}{l}\mathrm{Saima}\mathrm{wants}\mathrm{to}\mathrm{put}\mathrm{a}\mathrm{lace}\mathrm{on}\mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{a}\mathrm{circular}\mathrm{table}\\ \mathrm{cover}\mathrm{of}\mathrm{diameter}1.5\mathrm{m}.\mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{lace}\\ \mathrm{required}\mathrm{and}\mathrm{also}\mathrm{find}\mathrm{its}\mathrm{cost}\mathrm{if}\mathrm{one}\mathrm{meter}\mathrm{of}\mathrm{the}\mathrm{lace}\\ \mathrm{costs}\u20b9\; 15.\left(\mathrm{Take\pi}\hspace{0.33em}=\; 3.14\right)\end{array}$

**Ans**

\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\times \text{3}\text{.14}\times \frac{d}{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\times 3.14\frac{1.5}{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.71\text{\hspace{0.17em}}\text{m}\\ \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{Cost of 1 m of lace}=\text{\u20b9 15}\\ \text{Therefore cost of 4}\text{.71 m of lace}=\text{\u20b915}\times \text{4}\text{.71}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\u20b970}\text{.65}\end{array}

**Q.7 ****Find the perimeter of the adjoining figure, which is a semicircle including its diameter.**

**Ans**

\begin{array}{l}\text{Radius}=\text{5 cm}\\ \text{Length of curved part}=\pi \text{r}\\ \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}}=\frac{22}{7}\times 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}}=15.71\text{cm}\\ \text{Total permieter}=\text{Length of the curved part+Length of diameter}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{15}\text{.71+10}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{25}\text{.71 cm}\end{array}

**Q.8 **

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{polishing}\mathrm{a}\mathrm{circular}\mathrm{table}\u2013\mathrm{top}\mathrm{of}\mathrm{diameter}\\ 1.6\mathrm{m},\mathrm{if}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{polishing}\mathrm{is}\mathrm{Rs}15/{\mathrm{m}}^{\mathrm{2}}\mathrm{.}\left(\mathrm{Take\pi}\mathrm{=\; 3.14}\right)\end{array}$

**Ans**

\begin{array}{l}\text{Diameter}=\text{1}\text{.6 m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Radius}=\frac{1.6}{2}=0.8\text{m}\\ \text{Area}=\text{3}\text{.14}\times \text{0}\text{.8}\times \text{0}\text{.8}\\ \text{}\text{\hspace{0.17em}}=\text{2}{\text{.0096 m}}^{2}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Cost for polishing 1 m2 area}=\text{\u20b915}\\ \text{Cost for polishing 2}\text{.0096 m2 area}=\text{\u20b915}\times \text{2}\text{.0096}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\u20b930}\text{.14}\\ \text{So, it will cost \u20b930}\text{.14 for polishing such circular table}\text{.}\end{array}

**Q.9 **

$\begin{array}{l}\mathrm{Shazli}\mathrm{took}\mathrm{a}\mathrm{wire}\mathrm{of}\mathrm{length}44\mathrm{cm}\mathrm{and}\mathrm{bent}\mathrm{it}\mathrm{into}\mathrm{the}\\ \mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{circle}.\mathrm{Find}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{that}\mathrm{circle}.\mathrm{Also}\mathrm{find}\\ \mathrm{its}\mathrm{area}.\mathrm{If}\mathrm{the}\mathrm{same}\mathrm{wire}\mathrm{is}\mathrm{bent}\mathrm{into}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\\ \mathrm{square},\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{each}\mathrm{of}\mathrm{its}\mathrm{sides}?\\ \mathrm{Which}\mathrm{figure}\mathrm{encloses}\mathrm{more}.\end{array}$

**Ans**

\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}=\text{44 cm}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\times \frac{22}{7}\times r=44\\ \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}}r=7\text{cm}\end{array}

\begin{array}{l}\text{Area}=\pi {\text{r}}^{2}\\ \text{}\text{\hspace{0.17em}}=\frac{22}{7}\times 7\times 7\\ \text{}\text{\hspace{0.17em}}=154{\text{cm}}^{2}\\ \text{If the wire is bent into a square, then the length of each}\\ \text{side would be}=\frac{44}{4}=11\text{cm}\\ \text{Area of square}={\left(11\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=121{\text{cm}}^{2}\\ \text{Therefore, circle encloses more area}\text{.}\end{array}

**Q.10 **

$\begin{array}{l}\mathrm{From}\mathrm{a}\mathrm{circular}\mathrm{card}\mathrm{sheet}\mathrm{of}\mathrm{radius}14\mathrm{cm},\mathrm{two}\mathrm{circles}\mathrm{of}\\ \mathrm{radius}3.5\mathrm{cm}\mathrm{and}\mathrm{a}\mathrm{rectangle}\mathrm{of}\mathrm{length}3\mathrm{cm}\mathrm{and}\mathrm{breadth}\\ 1\mathrm{cm}\mathrm{are}\mathrm{removed}.\; (\mathrm{as}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{adjoining\; figure}).\\ \mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{sheet}.\left(\mathrm{Take\pi}\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}$

**Ans**

\begin{array}{l}\text{Area of bigger cirle}=\frac{22}{7}\times 14\times 14\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=616{\text{cm}}^{2}\\ \text{Area of 2 small circles}=\text{2}\times \pi {\text{r}}^{2}\\ \text{}\text{}\text{}\text{}\text{}=2\times \frac{22}{7}\times 3.5\times 3.5\\ \text{}\text{}\text{}\text{}\text{}=77{\text{cm}}^{2}\\ \text{Area of rectangle}=\text{Length}\times \text{Breadth}\\ \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{3}\times \text{1}={\text{3 cm}}^{2}\\ \text{Remaining area of sheet}={\text{616 cm}}^{2}-{\text{77 cm}}^{2}-{\text{3 cm}}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{536 cm}}^{2}\end{array}

**Q.11 **

$\begin{array}{l}\mathrm{A}\mathrm{circle}\mathrm{of}\mathrm{radius}2\mathrm{cm}\mathrm{is}\mathrm{cut}\mathrm{out}\mathrm{from}\mathrm{a}\mathrm{square}\mathrm{piece}\mathrm{of}\mathrm{an}\\ \mathrm{aluminium}\mathrm{sheet}\mathrm{of}\mathrm{side}6\mathrm{cm}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{left}\\ \mathrm{over}\mathrm{aluminium}\mathrm{sheet}?\left(\mathrm{Take}\mathrm{\pi}=3.14\right)\end{array}$

**Ans**

\begin{array}{l}\text{Area of square-shaped sheet}={\left(\text{side}\right)}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(6{\text{cm}}^{2}\right)}^{2}=36\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{Area of circle}=\text{3}\text{.14}\times 2\times 2\\ \text{}\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}}=12.56{\text{cm}}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Remaining area}={\text{36 cm}}^{\text{2}}-\text{12}{\text{.56 cm}}^{\text{2}}\\ \text{}\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{\hspace{0.17em}}=\text{23}{\text{.44 cm}}^{\text{2}}\end{array}

**Q.12 ****The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? (Take π = 3.14)**

**Ans**

\begin{array}{l}\text{Circumference}=\text{2}\pi r=31.4\text{\hspace{0.17em}}\text{cm}\\ \text{}\text{2}\times \text{3}\text{.14}\times \text{r}=\text{31}\text{.4}\\ \text{}\text{}\text{}\text{r}=\frac{31.4}{2\times 3.14}\\ \text{}\text{}\text{}=5\text{cm}\\ \text{}\text{\hspace{0.17em}}\text{So, Area}=\text{3}\text{.14}\times \text{5}\times \text{5}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}=\text{78}{\text{.50 cm}}^{2}\end{array}

**Q.13 ****A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)**

**Ans**

\begin{array}{l}\text{Radius of the flower bed}=\frac{66}{2}=33\text{\hspace{0.17em}}\text{m}\\ \text{So, radius of flower bed and path together}=\text{33+4}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{37m}\\ \text{Area of flower baed and path together}=\text{3}\text{.14}\times 37\times 37\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4298.66{\text{m}}^{2}\\ \text{Area of flower bed}=\text{3}\text{.14}\times \text{33}\times \text{33}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{3419}{\text{.46 m}}^{2}\\ \text{Area of path}=\text{4298}{\text{.66 m}}^{\text{2}}-\text{3419}{\text{.46m}}^{\text{2}}\\ \text{}\text{}\text{}=\text{879}{\text{.20m}}^{2}\end{array}

**Q.14 **

$\begin{array}{l}\mathrm{A}\mathrm{circular}\mathrm{flower}\mathrm{garden}\mathrm{has}\mathrm{an}\mathrm{area}\mathrm{of}314{\mathrm{m}}^{\mathrm{2}}.\mathrm{A}\mathrm{sprinkle\; r}\\ \mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{garden}\mathrm{can}\mathrm{cover}\mathrm{an}\mathrm{area}\mathrm{that}\mathrm{has}\mathrm{a}\\ \mathrm{radius}\mathrm{of}12\mathrm{m}.\mathrm{Will}\mathrm{the}\mathrm{sprinkler}\mathrm{water}\mathrm{the}\mathrm{entire}\mathrm{garden}?\\ \left(\mathrm{Take\pi}=3.14\right)\end{array}$

**Ans**

\begin{array}{l}\text{Area}=\pi {\text{r}}^{2}=314{\text{m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}3.14\times {r}^{2}=314\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}=100\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=10\text{m}\\ \text{Yes, the sprinkle will water the whole garden}\text{.}\end{array}

**Q.15 ****Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)**

**Ans**

\begin{array}{l}\text{Radius of outer circle}=\text{19 m}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Circumference}=\text{2}\pi \text{r}\\ \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{2}\times \text{3}\text{.14}\times \text{19}\\ \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{119}\text{.32 m}\\ \text{Radius of the inner circle}=\text{19}-\text{10}=\text{9 m}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Circumference}=\text{2}\pi \text{r}\\ \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{2}\times \text{3}\text{.14}\times \text{9}\\ \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{56}\text{.52 m}\end{array}

**Q.16 ****How many times a wheel of radius 28 cm must rotate to go 352 m? (take π****=22/7)**

**Ans**

\text{Radius, r}=\text{28 cm}

\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\times \frac{22}{7}\times 28\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=176\text{cm}\\ \text{Number of rotatinos}=\frac{\text{Total distance to be covered}}{\text{Circumference of the wheel}}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{352\text{m}}{176\text{\hspace{0.17em}}\text{cm}}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{35200}{176}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=200\\ \text{Therefore, it will rotate 200 times}\text{.}\end{array}

**Q.17 ****The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)**

**Ans**

\begin{array}{l}\text{Since, Distance travelled by the tip of minute hand}\\ \text{}=\text{Circumference of the clock}\\ \text{}=\text{2}\pi \text{r}\\ \text{}=\text{2}\times \text{3}\text{.14}\times 1\text{5}\\ \text{}=\text{94}\text{.2 cm}\\ \text{Therfore, minute hand move 94}\text{.2 cm in 1 hour}\text{.}\end{array}