NCERT Solutions Class 7 Maths Chapter 11

Class 7 introduces students to several new Mathematics topics and concepts, which will be further built upon in higher grades. This makes it very important that Class 7 students focus on building a strong foundation in this subject at this stage.

Students can use the NCERT Solutions for Class 7 Mathematics Chapter 11 by Extramarks for their preparations. They can get detailed and accurate solutions to questions given in the NCERT textbook, which will help them understand how to solve the different kinds of problems in a step-by-step manner. If they ever get stuck on a question, they can always refer to the solutions prepared by subject matter experts.

NCERT Solutions Class 7 Maths Chapter 11

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

Chapter 11 – Perimeter and Area Exercise
Exercise 11.1	Questions & Solutions
Exercise 11.2	Questions & Solutions
Exercise 11.3	Questions & Solutions
Exercise 11.4	Questions & Solutions

NCERT Solutions for Class 7 Maths Chapter 11 –

NCERT Class 7 Mathematics Chapter 11 teaches students how to calculate the perimeters and areas of different types of basic shapes, including squares, circles, triangles and parallelograms. Students are introduced to a lot of different formulas that they must remember and retain to perform well in the exam. They are also taught about how to handle the conversion of measurement units.

Some of the core concepts and formulas included in the chapter are as follows:

Perimeter is the total length of the boundary of a closed rectilinear figure.
The surface circled by a two-dimensional figure is its area.
Area of a Parallelogram = Base × Altitude
Area of a Triangle = ½ × Base × Altitude corresponding to the base
Circumference of a circle = 2πr.
Area of a Circle = πr²
1 cm² = 100 mm²
1m² = 10000 cm²
1 hectare = 10000 m²
The perimeter of a regular polygon = Length of a side × Number of the sides

Triangle: A triangle is a plane figure surrounded by three lines segments.

Rectilinear Figure: A plane figure encompassed by line segments is called a rectilinear figure. None of the two sides of it intersects at a point other than the vertex.

Quadrilateral: It is a plane figure surrounded by four long segments.

Parallelogram: Parallelogram is a plane figure encircled by four long segments. The opposite sides of a parallelogram are equal and parallel in length.

Area of a Triangle: The area of a triangle is ½ × Base × Height (Altitude corresponding to the base).

Example: base = 6 cm, height = 12cm. What is the area of a triangle?

As the formula goes; ½ × Base × Height

= (½ × 6 × 12) cm² = 36 cm²

Area of a Parallelogram: Base × Altitude (height)

Example: base = 10 cm and height = 5 cm

The formula of parallelogram: base × height

= (10 × 5) cm²

= 50 cm²

Circumference of a Circle: Perimeter of a circle is similar to the circumference of a circle. The students would draw four circles of different radii and find their circumference with the ratio of the circumferences to the diameter. The ratio will be more than three times its diameter.

Area of a Circle: The area a circle occupies on a flat plane which has its length and breadth, is known as the area of a surface. The formula to find the area is πr².

Conversion of Units

Units of Length	Units of Area
1 cm = 10 mm	1 cm² = (10 x 10) mm² = 100 mm²
1 dm = 10 cm	1 dm² = (10 x 10) cm² = 100 cm²
1 m = 10 dm	1 m² = (10 x 10) dm² = 100 dm²
1 dam = 10 m	1 dam² = (10 x 10) m² = 100 m²
1 hm = 10 dam	1 hm² = ( 10 x 10 ) dam² = 100 dam²
1 km = 10 hm	1 km² = ( 10 x 10) hm² = 100 hm²

Note:

1m = 100 cm ∴ 1 m² = 10000 cm²

1 km = 1000 m ∴ 1 km² = 10, 00, 000 m²

1 hectare (ha) = 100 m x 100m = 10000 m²

NCERT Solutions for Class 7 Maths

Mathematics is that one subject that many students struggle with. This is because getting good at Mathematics requires one to practise solving many problems and most students don’t do that. With NCERT Solutions of Class 7 Mathematics, Extramarks tries to provide students with a very valuable resource that they can use to get better at the subject and eventually score well in their final exams. Students can use these solutions if they ever get stuck on a problem or to learn the approach to how to solve the different kinds of problems encountered in any NCERT Mathematics Class 7 Chapter 11.

The NCERT Solutions by Extramarks are prepared by subject matter experts who give special attention to providing detailed and accurate solutions in a step-by-step manner so that students can easily grasp the problem-solving approach. Students can use these resources for their preparations, last-minute revisions, and for help with their assignments.

NCERT Solutions for Class 7

Apart from Mathematics, Extramarks provides comprehensive solutions to textbook questions for all of the other subjects covered in Class 7. All the solutions are prepared by their respective subject experts and extra care is given to providing accurate solutions which students can rely on. Students can download all these solutions from the Extramarks website.

NCERT Solutions Class 7 Maths Chapter-wise List

Chapter 1 – Integers

Chapter 2 – Fractions and Decimals

Chapter 3 – Data Handling

Chapter 4 – Simple Equations

Chapter 5 – Lines and Angles

Chapter 6 – The Triangle and Its Properties

Chapter 7 – Congruence of Triangles

Chapter 8 – Comparing Quantities

Chapter 9 – Rational Numbers

Chapter 10 – Practical Geometry

Chapter 11 – Perimeter and Area

Chapter 12 – Algebraic Expressions

Chapter 13 – Exponents and Powers

Chapter 14 – Symmetry

Chapter 15 – Visualising Solid Shapes

Q.1 Find the area of a square park whose perimeter is 320 m.

Ans

$\begin{array}{l} Let the side of the square park be ‘ a ‘ . \\ Since perimeter of a square = 4 a \\ So, we get \\ 4 a = 320 \\ a = \frac{320}{4} \\ = 80 m \\ Therefore, the area of the square park = a^{2} \\ = {(80 m)}^{2} \\ = 6400 m^{2} \end{array}$

Q.2 Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.

Ans

$\begin{array}{l} Let the breadth of the rectangular plot be x . \\ Area of the rectangular plot = length \times breadth \\ SO, we get \\ {440m}^{2} = 22 m \times x \\ x = \frac{440 m^{2}}{22 m} \\ = 20 m \\ Perimeter of the rectangular plot = 2 (length+breadth) \\ = 2 (20 m + 22 m) \\ = 2 (42 m) \\ = 84 m \end{array}$

Q.3 The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.

Ans

$\begin{array}{l} Let the breadth of the rectangular sheet be x . \\ Perimeter of the rectangular sheet = 2 (length+breadth) \\ 100 cm = 2 (35 cm + x) \\ \frac{100}{2} cm = (35 cm+x) \\ 50 cm = 35 cm+x \\ x = 50 cm - 35 cm \\ = 15 cm \\ Area of rectangular sheet = length \times breadth \\ = 35 cm \times 15 cm \\ = {525 cm}^{2} \end{array}$

Q.4

$\begin{array}{l} The area of a square park is the same as of a rectangular \\ park . If the side of the square park is 60 m and the length \\ of the rectangular park is 90 m, find the breadth of the \\ rectangular park . \end{array}$

Ans

$\begin{array}{l} Let the breadth of rectangular park be x . \\ Since, area of a square park is the same as of a rectangular \\ park . \\ Area of square = Area of rectangular park \\ {(60m)}^{2} = 90 m \times x \\ 3600 m^{2} = 90 m \times x \\ x = \frac{3600 m^{2}}{90 m} \\ = 40 m \end{array}$

Q.5

$\begin{array}{l} A wire is in the shape of a rectangle . Its length is 40 cm \\ and breadth is 22 cm . If the same wire is rebent in the \\ shape of a square, what will be the measure of each side . \\ Also find which shape encloses more area ? \end{array}$

Ans

$\begin{array}{l} Perimeter of a rectangle = Perimeter of square \\ 2 (length+breadth) = 4 \times side \\ 2 (40 m + 22 m) = 4 \times side \\ side = \frac{124}{4} \\ = 31 cm \\ S o, \\ Area of rectangle = length \times breadth \\ = 40cm \times 22cm \\ = {880cm}^{2} \\ Area of square = {(side)}^{2} \\ = {(31 cm)}^{2} \\ = 961 {cm}^{2} \\ Therefore, the square-shaped wire encloses more area . \end{array}$

Q.6

$\begin{array}{l} The perimeter of a rectangle is 130 cm . If the breadth \\ of the rectangle is 30 cm, find its length . Also find the \\ area of the rectangle . \end{array}$

Ans

$\begin{array}{l} Let the length of the rectangle be x cm . \\ Perimeter of rectangle = 2 (length + breadth) \\ 130 cm = 2(30 cm + x) \\ \frac{130}{2} cm = 30 cm + x \\ x = 65 cm - 30 cm \\ = 35 cm \\ Now, \\ area of rectangle = length \times breadth \\ = 30 cm \times 35 cm \\ = {1050 cm}^{2} \end{array}$

Q.7 Find the area of each of the following parallelogram :

Ans

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

Q.8 Find the area of each of the following triangles :

Ans

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

Q.9 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

$\begin{array}{l} Let the height be h and base be b . \\ (a) Area of parallelogram=Base \times Height \\ So,we get \\ {246 cm}^{2} = 20 cm \times h \\ h= \frac{246 {cm}^{2}}{20 cm} \\ = 12.3 cm \\ (b) Area of parallelogram=Base \times Height \\ So,we get \\ 154 {.5 cm}^{2} = b \times 15 cm \\ b= \frac{154.5 {cm}^{2}}{15 cm} \\ = 10.3 cm \end{array}$ $\begin{array}{l} (c) Area of parallelogram=Base \times Height \\ So,we get \\ 78 {.72 cm}^{2} = b \times 8 .4 cm \\ b= \frac{48.72 {cm}^{2}}{8.4 cm} \\ = 5.8 cm \\ (d) Area of parallelogram=Base \times Height \\ So,we get \\ 16 {.38 cm}^{2} = 15 .6 cm \times h \\ h= \frac{16.38 {cm}^{2}}{15.6 cm} \\ = 1.05 cm \\ So, we get \\ \begin{matrix} S .No . & Base & Height & Area of the Parallelogram \\ a . & 20 cm & 12.3 cm & 246 {cm}^{2} \\ b . & 10.3 cm & 15 cm & 154.5 {cm}^{2} \\ c . & 5.8 cm & 8.4 cm & 78.72 {cm}^{2} \\ d . & 15.6 cm & 1.05 cm & 16.38 {cm}^{2} \end{matrix} \end{array}$

Q.10 Find the missing values :

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

Ans

$\begin{array}{l} (a) Let the height be h and base be b . \\ Area of triangle = \frac{1}{2} \times base \times height \\ {87cm}^{2} = \frac{1}{2} \times 15 cm \times h \\ h = \frac{87 {cm}^{2} \times 2}{15 cm} \\ = 11.6 cm \\ (b) Let the height be h and base be b . \\ Area of triangle = \frac{1}{2} \times base \times height \\ {1256 mm}^{2} = \frac{1}{2} \times b \times 31 .4 mm \\ b = \frac{1256 {mm}^{2} \times 2}{31.4 mm} \\ = 80 mm \end{array}$ $\begin{array}{l} (c) Let the height be h and base be b . \\ Area of triangle = \frac{1}{2} \times base \times height \\ 170 {.5 cm}^{2} = \frac{1}{2} \times 22 cm \times h \\ h = \frac{170.5 {cm}^{2} \times 2}{22 cm} \\ = 15.5 cm \\ So we get \\ \begin{matrix} Base & Height & Area of Triangle \\ 15 cm & 11.6 cm & 84 {cm}^{2} \\ 80 mm & 31.4 mm & 1256 {mm}^{2} \\ 22 cm & 15.5 cm & 170.5 {cm}^{2} \end{matrix} \end{array}$

Q.11

$\begin{array}{l} PQRS is a parallelogram . QM is the height from Q to SR \\ and QN is the height from Q to PS . If SR =12 cm and \\ QM = 7.6 cm . Find : \\ (a) the area of the parallegram PQRS \\ (b) QN, if PS = 8 cm \end{array}$

Ans

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

Q.12

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

Ans

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

Q.13

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

Ans

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

Q.14

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

Ans

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

Q.15

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

Ans

$\begin{array}{l} (a) \\ Circumference of a circle is = 2 π r \\ = 2 \times \frac{22}{7} \times 14 \\ = 2 \times 22 \times 2 \\ = 88 cm \\ (b) \\ Circumference of a circle is = 2 π r \\ = 2 \times \frac{22}{7} \times 28 \\ = 2 \times 22 \times 4 \\ = 176 mm \end{array}$ $\begin{array}{l} (c) \\ Circumference of a circle is = 2 π r \\ = 2 \times \frac{22}{7} \times 21 \\ = 2 \times 22 \times 3 \\ = 132 cm \end{array}$

Q.16

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

Ans

$\begin{array}{l} (a) \\ Area of a circle = π r^{2} \\ = \frac{22}{7} {(14)}^{2} \\ = 616 {mm}^{2} \end{array}$ $\begin{array}{l} (b) \\ Radius = \frac{Diameter}{2} = \frac{49}{2} = 24.5 mm \\ Area of a circle = π r^{2} \\ = \frac{22}{7} {(24.5)}^{2} \\ = 1886.5 m^{2} \\ (c) \\ Area of a circle = π r^{2} \\ = \frac{22}{7} {(5)}^{2} \\ = \frac{550}{7} {cm}^{2} \end{array}$

Q.17 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =

$\frac{22}{7}$

)

Ans

$\begin{array}{l} Since, circumference = 2 π r \\ So, we get \end{array}$ $\begin{array}{l} 154 = 2 \times \frac{22}{7} \times r \\ r = \frac{154 \times 7}{2 \times 22} \\ = \frac{49}{2} m = 4.5 m \\ Now, \\ Area = π r^{2} \\ = \frac{22}{7} \times \frac{49}{2} \times \frac{49}{2} \\ = 1886.5 m^{2} \end{array}$

Q.18 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

$\begin{array}{l} Outer radius of circular sheet = 4 cm \\ Inner radius of circular sheet = 3 cm \end{array}$ $\begin{array}{l} Remaining area = - (3.14 \times 3 \times 3) \\ = 50.24 - 28.26 \\ = 21.98 {cm}^{2} \end{array}$

Q.19 Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Ans

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

Q.20

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

Ans

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

Q.21

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

Ans

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

Q.22

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

Ans

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

Q.23

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

Ans

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

Q.24 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

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

Q.25

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

Ans

$\begin{array}{l} Area = π r^{2} = 314 m^{2} \\ 3.14 \times r^{2} = 314 \\ r^{2} = 100 \\ r = 10 m \\ Yes, the sprinkle will water the whole garden . \end{array}$

Q.26 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Ans

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

Q.27 How many times a wheel of radius 28 cm must rotate to go 352 m? (take π =

$\frac{22}{7}$

)

Ans

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

Q.28 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

$\begin{array}{l} Since, Distance travelled by the tip of minute hand \\ = Circumference of the clock \\ = 2 π r \\ = 2 \times 3 .14 \times 15 \\ = 94 .2 cm \\ Therfore, minute hand move 94 .2 cm in 1 hour . \end{array}$

Q.29 A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Ans

$\begin{array}{l} Length (L) of garden = 90 m \\ Breadth (B) of garden = 75 m \\ Area of garden = L \times B \\ = 90 m \times 75 m \\ = {6750 m}^{2} \end{array}$ $\begin{array}{l} From the figure, it can be observed that the new length \\ and breadth of the garden,when path is also included are \\ 100 m and 85 m respectively \\ Area of the garden including path = 100 m \times 80 m \\ = {8000 m}^{2} \\ Area of the path \\ = Area of the garden including path - Area of garden \\ = 8000 m^{2} - {6750 m}^{2} \\ = 1750 m^{2} \\ 1 hectare = {10000 m}^{2} \\ Therefore, \\ area of the garden in hectare = \frac{6750}{10000} \\ = 0.675 hectare \end{array}$

Q.30 A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65m. Find the area of the path.

Ans

$\begin{array}{l} Length (L) of park = 125 m \\ Breadth (B) of park = 65 m \\ Area of park = L \times B \\ = 125 m \times 65 m \\ = {8125 m}^{2} \end{array}$ $\begin{array}{l} From the figure, it can be observed that the new length \\ and breadth of the park,when path is also included are \\ 131 m and 71 m respectively \\ Area of the park including path = 131 m \times 71 m \\ = {9301 m}^{2} \\ Area of the path \\ = Area of the park including path - Area of park \\ = 9301 m^{2} - {8125 m}^{2} \\ = 1176 m^{2} \end{array}$

Q.31 A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Ans

$\begin{array}{l} Length (L) of cardboard = 125 m \\ Breadth (B) of cardboard = 65 m \\ Area of cardboard including margin = L \times B \\ = 8 cm \times 5 cm \\ = 40 {cm}^{2} \end{array}$ $\begin{array}{l} From the figure, it can be observed that the new length \\ and breadth of the cardboard,when margin is not included \\ are 5 cm and 2 cm respectively \\ Area of the cardboard not including margin = 5 cm \times 2 m \\ = {10 cm}^{2} \\ Area of the margin \\ = Area of the cardboard including margin \\ - Area of cardboard not including margin \\ = 40 {cm}^{2} - {10 cm}^{2} \\ = 30 {cm}^{2} \end{array}$

Q.32

$\begin{array}{l} Two cross roads, each of width 10 m, cut at right angles \\ through the centre of a rectangular park of length 700 m \\ and breadth 300 m and parallel to its sides . Find the area \\ of the roads . Also find the area of the park excluding cross \\ roads . Give the answer in hectares . \end{array}$

Ans

$\begin{array}{l} Length (L) of the park = 700 m \\ Breadth (B) of the park = 300 m \\ Area of park = 700 \times 300 \\ = {210000 m}^{2} \\ Length of road PQRS = 700 m \\ Length of road ABCD = 300 m \\ Width of each road = 10 m \\ Area of the roads = area(PQRS)+area(ABCD) - area(KLMN) \\ = (700 \times 10) + (300 \times 10) - (10 \times 10) \\ = 7000 + 3000 - 100 \\ = 9900 m^{2} \\ = 0.99 hectare \end{array}$ $\begin{array}{l} Area of park excluding roads = 210000-9900 \\ = {200100 m}^{2} \\ = 20.01 hectare \end{array}$

Q.33 Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any left? (π = 3.14)

Ans

$\begin{array}{l} Perimeter of the square = 4 \times side of the square \\ = 4 \times 4 \\ = 16 cm \\ Perimeter of circular pip = 2 π r \\ = 3 \times 3 .14 \times 4 \\ = 25 .12 cm \\ Therefore, \\ length of chord left with Pragya = 25 .12 cm - 16 cm \\ = 9 .12 cm \end{array}$

Q.34 The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed.
(iv) the circumference of the flower bed.

Ans

$\begin{array}{l} (i) Area of whole land = Length \times Breadth \\ = 10 \times 5 \\ = {50 m}^{2} \\ (ii) Area of flower bed = π r^{2} \\ = 3.14 \times 2 \times 2 \\ = 12.56 m^{2} \\ (iii) Area of the lawn excluding the flower bed \\ = Area of whole land - Area of flower bed \\ = 50-12 .56 \\ = 37 {.44 m}^{2} \end{array}$

Q.35 In the following figures, find the area of the shaded portions:

Ans

$\begin{array}{l} (i) \\ Area of EFDC = area(ABCD) - area(BCE) - area(AFE) \\ = (18 \times 10) - \frac{1}{2} (10 \times 8) - \frac{1}{2} (6 \times 10) \\ = 180 - 40 - 30 \\ = 110 {cm}^{2} \\ (ii) \\ area(QTU) = area(PQRS) - area(TSU) - area(RUQ) - area(PQT) \\ = (20 \times 20) - \frac{1}{2} (10 \times 10) - \frac{1}{2} (20 \times 10) - \frac{1}{2} (20 \times 10) \\ = 400 - 50 - 100 - 100 \\ = 150 {cm}^{2} \end{array}$

Q.36

$\begin{array}{l} Find the area of the quadrilateral ABCD . \\ Here, AC =22 cm, BM =3 cm, DN =3 cm, \\ and BM ⊥ AC, DN ⊥ AC . \end{array}$

Ans

$\begin{array}{l} area(ABCD) = area(ABC)+area(ADC) \\ = \frac{1}{2} (3 \times 22) \frac{1}{2} (3 \times 22) \\ = 33 + 33 \\ = 66 {cm}^{2} \end{array}$

Q.37

$\begin{array}{l} The length and the breadth of a rectangular piece of land \\ are 500 m and 300 m respectively . Find \\ (i) its area \\ (ii) the cost of the land, if 1 m^{2} of the land costs ₹ 10,000. \end{array}$

Ans

$\begin{array}{l} (i) \\ Area = Length \times breadth \\ = 500 m \times 3000 \times m \\ = {150000m}^{2} \\ (ii) \\ {1 m}^{2} land cost = ₹ 10,000 \\ {So, cost of 150000 m}^{2} land = ₹ 10,000 \times 150000 \\ = ₹ 1,500,000,000 \end{array}$

Q.38

$\begin{array}{l} A door of length 2 m and breadth 1 m is fitted in a wall . \\ The length of the wall is 4.5 m and the breadth is 3.6 m . \\ Find the cost of white washing the wall, if the rate of \\ white washing the wall is ₹ 20 per m^{2} . \end{array}$

Ans

$\begin{array}{l} Area of the wall = 4 .5 cm \times 3 .6 cm \\ = 16 {.2 m}^{2} \\ Area of the door = 2 \times 1 \\ = {2m}^{2} \\ Area to be white-washed = 16 {.2 m}^{2} - {2 m}^{2} \\ = 14 {.2 m}^{2} \\ {Cost of white-washing 1m}^{2} area = ₹20 \\ So, cost of white-washing 14 .2m2 area = 14 .2 \times 20 \\ = ₹ 284 \end{array}$

Q.39

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

Ans

$\begin{array}{l} Given, d = 21 m \\ So, r = \frac{21}{2} m \\ Circumference = 2 π r \\ = 2 \times \frac{22}{7} \times \frac{21}{2} \\ = 66 m \\ Length of the rope required for fencing = 2 \times 66 m \\ = 132 m \\ Cost of 1 m rope = ₹4 \\ So, cost of 132 m rope = ₹ 4 \times 132 \\ = ₹528 \end{array}$

Q.40

$\begin{array}{l} Saima wants to put a lace on the edge of a circular table \\ cover of diameter 1.5 m . Find the length of the lace \\ required and also find its cost if one meter of the lace \\ costs ₹ 15. (Takeπ = 3.14) \end{array}$

Ans

$\begin{array}{l} Circumference = 2 π r \\ = 2 \times 3 .14 \times \frac{d}{2} \\ = 2 \times 3.14 \frac{1.5}{2} \\ = 4.71 m \\ Cost of 1 m of lace = ₹ 15 \\ Therefore cost of 4 .71 m of lace = ₹15 \times 4 .71 \\ = ₹70 .65 \end{array}$

Q.41

$\begin{array}{l} Find the cost of polishing a circular table - top of diameter \\ 1.6 m, if the rate of polishing is Rs 15/ m^{2} . (Takeπ = 3.14) \end{array}$

Ans

$\begin{array}{l} Diameter = 1 .6 m \\ Radius = \frac{1.6}{2} = 0.8 m \\ Area = 3 .14 \times 0 .8 \times 0 .8 \\ = 2 {.0096 m}^{2} \\ Cost for polishing 1 m2 area = ₹15 \\ Cost for polishing 2 .0096 m2 area = ₹15 \times 2 .0096 \\ = ₹30 .14 \\ So, it will cost ₹30 .14 for polishing such circular table . \end{array}$

Q.42

$\begin{array}{l} A verandah of width 2.25 m is constructed all along \\ outside a room which is 5.5 mlong and 4 m wide . Find : \\ (i) the area of the verandah . \\ (ii) the cost of cementing the floor of the verandah at the \\ rate of Rs 200 per m^{2} \end{array}$

Ans

$\begin{array}{l} (i) \\ Length (L) of room = 5 .5 m \\ Breadth (B) of room = 4 m \\ Area of room including margin = L \times B \\ = 5 .5 m \times 4 m \\ = 22 m^{2} \end{array}$ $\begin{array}{l} From the figure, it can be observed that the new length \\ and breadth of the room,when verandah is also included \\ are 10 m and 8 .5 m respectively \\ Area of the verandah including margin = 10 m \times 8 .5 m \\ = {85 m}^{2} \\ Area of the verandah \\ = Area of the room including verandah - Area of the room \\ = 85 m^{2} - {22 m}^{2} \\ = 63 m^{2} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} (ii) \\ {Cost of cementing 1m}^{2} area of floor of verandah = ₹ 200 \\ {Cost of cementing 63m}^{2} area of floor of verandah = ₹ 200 \times 63 \\ = ₹ 12600 \end{array}$

Q.43

$\begin{array}{l} A path 1 m wide is built along the border and inside a \\ square garden of side 30 m . Find : \\ (i) the area of the path \\ (ii) the cost of planting grass in the remaining portion of the \\ garden at the rate of ₹ 40 per m^{2} . \end{array}$

Ans

$\begin{array}{l} (i) \\ Side (a) of square garden = 30 m \\ Area of square garden including path = a^{2} \\ = {(30 m)}^{2} \\ = {900 m}^{2} \\ From the figure, it can be observed that the side of the square \\ garden, when path is not included, is 28 m . \\ Area of the square garden not including the path = {(28)}^{2} \\ = {784 m}^{2} \\ Area of path \\ = Area of the square garden including the path \\ - Area of square garden not including the path \\ = 900 m^{2} - 784 m^{2} \\ = 116 m^{2} \\ (ii) \\ {Cost of planting grass in 1 m}^{2} area of the garden = ₹40 \\ {Cost of planting grass in 784 m}^{2} area of the garden = ₹40 \times 784 \\ = ₹31360 \end{array}$

Q.44

$\begin{array}{l} Through a rectangular field of length 90 m and breadth \\ 60 m, two roads are constructed which are parallel to the \\ sides and cut each other at right angles through the centre \\ of the fields . If the width of each road is 3 m, find \\ (i) the area covered by the roads . \\ (ii) the cost of constructing the roads at the rate of \\ ₹ 110 per m^{2} . \end{array}$

Ans

$\begin{array}{l} Length (L) of the field = 90 m \\ Breadth (B) of the field = 60 m \\ Area of field = 90 \times 60 \\ = {5400 m}^{2} \\ Length of road PQRS = 90 m \\ Length of road ABCD = 60 m \\ Width of each road = 3 m \\ Area of the roads = area(PQRS)+area(ABCD) - area(KLMN) \\ = (90 \times 3) + (60 \times 3) - (3 \times 3) \\ = 270 + 180 - 100 \\ = 441 m^{2} \\ {Cost for constructing 1 m}^{2} road = ₹110 \\ {Cost for constructing 4411 m}^{2} road = ₹110 \times 441 \\ = ₹ 48510 \end{array}$

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FAQs (Frequently Asked Questions)

1. Where will I find the class 7 Mathematics chapter-wise solutions?

You will find the class 7 Mathematics chapter-wise solutions on the Extramarks website.

2. Why choose Extramarks NCERT solutions?

The primary reasons to choose Extramarks NCERT solutions are that they are:

drafted by experienced faculties
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3. What is chapter 11 of class 7 about?

Chapter 11 of Class 7 is about Area and Perimeter. Perimeter is the distance around a close figure whereas area is a metric to quantify the region occupied by a figure. Students will learn how to calculate the perimeters and areas of many different shapes including triangles, circles, squares, and parallelograms in this chapter.

4. What are the area and perimeter?

Perimeter is the distance that surrounds a specific area whereas, an area is the part of a plane that is encircled by the perimeter. An area is a particular shape that includes a square, rectangle, triangle, etc.

5. What is the formula for the area of a circle?

The formula of the area of a circle is πr2.

NCERT Solutions Class 7 Maths Chapter 11