NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area: If you want to cover the floor of the room. The first question that comes in mind is how many square meters of flooring you need? It can be done by measuring the area of the room. In the practical case, you will buy slightly higher than the required area since there may be cuts and joints required for the flooring. In CBSE NCERT solutions for class 7 maths chapter 11 perimeter and area, you will get questions related to applications of perimeter and area. Perimeter is the distance around a closed figure and the area is the space occupied by a close figure. The perimeter can be calculated by adding the length of all sides of the closed figure. In this chapter, you will study the perimeters of some simple geometry like rectangle, triangle, parallelogram, and circles. Also, you will learn to calculate the area of rectangle, triangle, parallelogram, and circles. In solutions of NCERT for class 7 maths chapter 11 perimeter and area, you will get all questions related to calculating the area and perimeter only. You should try to solve all the problems including examples for a better understanding of the concepts. Before solving the problems, you must convert all the given values to a common unit so you can get rid of the confusion. If you are given a complex geometry to calculate the area, you can divide it into smaller parts and the total area would be the sum of the area of all the parts. In CBSE NCERT solutions for class 7 maths chapter 11 perimeter and area, you will get some complex geometry problems which will give you more clarity. You can get NCERT solutions from class 6 to 12 by clicking on the above link. Here ypou will get solutions to four exercises of this chapter.

Exercise:11.1

Exercise:11.2

Exercise:11.3

Exercise:11.4

The main topics of the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area are:

11.1 Introduction

11.2 Squares And Rectangles

11.2.1 Triangles As Parts Of Rectangles

11.2.2 Generalising For Other Congruent Parts Of Rectangles

11.3 Area Of A Parallelogram

11.4 Area Of A Triangle

11.5 Circles

11.5.1 Circumference Of A Circle

11.5.2 Area Of Circle

11.6 Conversion Of Units

11.7 Applications

Solutions of NCERT for class 7 maths chapter 11 perimeter and area topic 11.2 squares and rectangles

How much space does a blackboard occupy?

The space of the board includes whole area of the board

What is the length of a wire required to fence a rectangular flower bed?

The length of the wire to fence a flower bed is the circumference of the flower bed

What distance would you cover by taking two rounds of a triangular park?

Distance covered by taking round a triangular park is equal to the circumference of the triangular park

How much plastic sheet do you need to cover a rectangular swimming pool?

plastic sheet need to cover is the area of the rectangular swimming pool

A square of side 1m has perimeter 4 m and area 1 m2. When all sides are increased by 1m then perimeter =8m and area= 4 m2. Similarly if we increase the length from 6m to 9m and breadth from 3m to 6m  of a rectangular the perimeter and area will increase

Area of a rectangle with length = 20cm, breadth = 5 cm is 100cm2 and perimeter is 50 cm. A rectangle with sides 50 cm and 2 cm ha area = 100cm2  but perimeter is 104 cm

CBSE NCERT solutions for class 7 maths chapter 11 perimeter and area-Exercise: 11.1

It is given that the length and the breadth of a rectangular piece of land are $500\; m$ and $300\; m$

Now, we know that

Area of the rectangle (A) = $l \times b = 500\times 300 = 150000 \ m ^2$

Therefore, area of  rectangular piece of land is $150000 \ m ^2$

It is given that the length and the breadth of a rectangular piece of land are $500\; m$ and $300\; m$

Now, we know that

Area of rectangle (A) = $l \times b = 500\times 300 = 150000 \ m ^2$

Now, it is given that  $1\; m^{2}$ of the land costs  $Rs.10.000$

Therefore,  cost of $150000 \ m ^2$  of land is $=10,000\times 150000= 1,500,000,000 \ Rs$

It is given that the perimeter of the square park is $320\; m$

Now, we know that

The perimeter of a square is (P) $= 4a$   , where a is the side of the square

$\Rightarrow 4a = 320$

$\Rightarrow a = \frac{320}{4}=80 \ m$

Now,

Area of the square (A) $= a^2$

$\Rightarrow a^2 = (80)^2 = 6400 \ m^2$

Therefore, the area of a square park  is  $6400 \ m^2$

It is given that the area of rectangular land is $440\; m^{2}$ and the length is $22\; m$

Now, we know that

Area of rectangle is $= length \times breath$

$\Rightarrow 440 = 22 \times breath$

$\Rightarrow breath = \frac{440}{22} = 20 \ m$

Now,

The perimeter of the rectangle is $=2(l+b)$

$\Rightarrow 2(l+b) = 2(20+22)=2\times 44 = 88 \ m$

Therefore, breath and perimeter of the rectangle are 20m and 88m respectively

It is given that  perimeter of a rectangular sheet is $100\; cm$ and length is $35\; cm$

Now, we know that

The perimeter of the rectangle is $=2(l+b)$

$\Rightarrow 100 = 2(l+b)$

$\Rightarrow 100 = 2(35+b)$

$\Rightarrow 2b = 100-70$

$\Rightarrow 2 = \frac{30}{2}=15 \ cm$

Now,

Area of rectangle is

$\Rightarrow l \times b = 35 \times 15 = 525 \ cm^2$

Therefore, breath and area  of the rectangle are 15cm and $525 \ cm^2$ respectively

It is given that the area of a square park is the same as of a rectangular park and side of the square park is $60\; m$ and the length of the rectangular park is $90\; m$

Now, we know that

Area of square $= a^2$

Area of rectangle $=l \times b$

Area of square = Area of rectangle

$a^2=l \times b$

$\Rightarrow 90\times b = (60)^2$

$\Rightarrow b = \frac{3600}{90} = 40 \ m$

Therefore, breadth of the rectangle is 40 m

It is given that the length of rectangular wire is $40\; cm$ and breadth is $22\; cm$.

Now, if it reshaped into a square wire

Then,

The perimeter of rectangle = perimeter of the square

$2(l+b)= 4a$

$\Rightarrow 2(40+22) = 4a$

$\Rightarrow a = \frac{124}{4}= 31 \ cm$

Now,

Area of rectangle $= l \times b = 40 \times 22 = 880 \ cm^2$

Area of square $= a^2 = (31)^2 = 961 \ cm^2$

Therefore, the side of the square is 31 cm and we can clearly see that square-shaped wire encloses more area

It is given that the perimeter of a rectangle is $130\; cm$ and breadth is $30\; cm$

Now, we know that

The perimeter of the rectangle is $=2(l+b)$

$\Rightarrow 130 = 2(l+30)$

$\Rightarrow 2l = 130- 60$

$\Rightarrow l = \frac{70}{2} = 35 \ cm$
Now,

Area of rectangle is $= length \times breath$

$\Rightarrow 35 \times 30 = 1050 \ cm^2$

Therefore, the length and area of the rectangle are 35cm and $1050 \ cm^2$ respectively

It is given that the length of door is $2\; m$ and breadth is  $1\; m$ and the length of the wall is $4.5\; m$ and the breadth is$3.6\; m$

Now, we know that

Area of rectangle is $= length \times breath$

Thus, the area of the wall is

$\Rightarrow 4.5 \times 3.6 =16.2 \ m^2$

And

Area of the door is

$\Rightarrow 2 \times 1 =2 \ m^2$

Now, Area to be painted = Area of wall - Area of door = 16.2 - 2 = 14.2 $\ m^2$

Now, the cost of whitewashing the wall, at the rate of $Rs.20\; per\; m^{2}$ is

$\Rightarrow 14.2 \times 20 = 284 \ Rs$

Therefore, the cost of whitewashing the wall, at the rate of $Rs.20\; per\; m^{2}$ is Rs 284

NCERT solutions for class 7 maths chapter 11 perimeter and area topic 11.2.2 generalising for other congruent parts of rectangles

The total area of rectangle $=6\times 4=24cm^2$

i) The first rectangle is divided into 6 equal parts. So the area of each part will be one-sixth of the total area

$area= \frac{6\times4}{6}=4cm^2$

ii) The rectangle is divided into 4 equal parts, area of each part= one-forth area of rectangle= 6 cm2

iii) and (iv) are divided into two equal parts. Area of each part will be one half the total area of rectangle= 12  cm2

v) Area of rectangle is divided into 8 equal parts. Area of one part is one-eighth of the total area of rectangle = 3 cm2

Solutions of NCERT for class 7 maths chapter 11 perimeter and area topic 11.3 area of a parallelogram

Question:1(i) Find the area of following parallelograms:

Area of the parallelogram is the product of base and height

$area=8\times3.5=28cm^2$

Question:

Area of the parallelogram is the product of base and height

$area=8\times2.5=20cm^2$

CBSE NCERT solutions for class 7 maths chapter 11 perimeter and area-Exercise: 11.2

Question:

We know that

Area of parallelogram  $= base \times height$

Here,

Base of parallelogram = 7cm

and

Height of parallelogram = 4 cm

$\Rightarrow 7 \times 4 = 28 \ cm^2$

Therefore, the area of the parallelogram is  $28 \ cm^2$

Question:

We know that

Area of parallelogram  $= base \times height$

Here,

Base of parallelogram = 5cm

and

Height of parallelogram = 3 cm

$\Rightarrow 3 \times 5 = 15 \ cm^2$

Therefore, the area of the parallelogram is  $15 \ cm^2$

Question:

We know that

Area of parallelogram  $= base \times height$

Here,

Base of parallelogram = 2.5cm

and

Height of parallelogram = 3.5 cm

$\Rightarrow 3.5 \times 2.5 = 8.75 \ cm^2$

Therefore, area of parallelogram is  $8.75 \ cm^2$

Question:

We know that

Area of parallelogram  $= base \times height$

Here,

Base of parallelogram = 5cm

and

Height of parallelogram = 4.8 cm

$\Rightarrow 5 \times 4.8 = 24 \ cm^2$

Therefore, the area of a parallelogram is  $24 \ cm^2$

Question:

We know that

Area of parallelogram  $= base \times height$

Here,

Base of parallelogram = 2 cm

and

Height of parallelogram = 4.4 cm

$\Rightarrow 2 \times 4.4 = 8.8 \ cm^2$

Therefore, area of parallelogram is  $8.8 \ cm^2$

Question:

We know that

Area of triangle  $=\frac{1}{2}\times base \times height$

Here,

Base of triangle = 4 cm

and

Height of triangle =3 cm

$\Rightarrow \frac{1}{2} \times 4 \times 3 = 6 \ cm^2$

Therefore, area of triangle is  $6 \ cm^2$

Question:

We know that

Area of triangle  $=\frac{1}{2}\times base \times height$

Here,

Base of triangle = 5 cm

and

Height of triangle =3.2 cm

$\Rightarrow \frac{1}{2} \times 5 \times 3.2 = 8 \ cm^2$

Therefore, the area of the triangle is  $8 \ cm^2$

Question:

We know that

Area of triangle  $=\frac{1}{2}\times base \times height$

Here,

Base of triangle = 3 cm

and

Height of triangle =4 cm

$\Rightarrow \frac{1}{2} \times 3 \times 4 = 6 \ cm^2$

Therefore, area of triangle is  $6 \ cm^2$

Question:

We know that

Area of triangle  $=\frac{1}{2}\times base \times height$

Here,

Base of triangle = 3 cm

and

Height of triangle =2 cm

$\Rightarrow \frac{1}{2} \times 2 \times 3 = 3 \ cm^2$

Therefore, the area of the triangle is  $3 \ cm^2$

Question:

We know that

Area of parallelogram  $= base \times height$

a) Here, base and area of parallelogram is given

$\Rightarrow 246= 20 \times height$

$\Rightarrow height = \frac{246}{20}= 12.3 \ cm$

b) Here height and area of parallelogram is given

$\Rightarrow 154.5= 15 \times base$

$\Rightarrow base = \frac{154.5}{15}=10.3 \ cm$

c) Here height and area of parallelogram is given

$\Rightarrow 48.72= 8.4 \times base$

$\Rightarrow base = \frac{48.72}{8.4}=5.8 \ cm$

d) Here base and area of parallelogram is given

$\Rightarrow 16.38= 15.6 \times height$

$\Rightarrow height = \frac{16.38}{15.6}=1.05 \ cm$

Question:

We know that

Area of triangle  $= \frac{1}{2}\times base \times height$

a) Here, the base and area of the triangle is given

$\Rightarrow 87= \frac{1}{2} \times 15 \times height$

$\Rightarrow height = \frac{87}{7.5}= 11.6 \ cm$

b) Here height and area of the triangle is given

$\Rightarrow 1256= \frac{1}{2}\times 31.4 \times base$

$\Rightarrow base = \frac{1256}{15.7}=80 \ mm$

c) Here base and area of the triangle is given

$\Rightarrow 170.5= \frac{1}{2} \times 22 \times height$

$\Rightarrow height = \frac{170.5}{11}=15.5 \ cm$

$\inline QM=7.6 \; cm.$ Find:

We know that

Area of parallelogram  $= base \times height$

Here,

Base of parallelogram = 12 cm

and

Height of parallelogram = 7.6 cm

$\Rightarrow 12 \times 7.6 = 91.2 \ cm^2$

Therefore, area of parallelogram is  $91.2 \ cm^2$

$\inline QM=7.6 \; cm.$ Find:

We know that

Area of parallelogram  $= base \times height$

Here,

Base of parallelogram = 12 cm

and

Height of parallelogram = 7.6 cm

$\Rightarrow 12 \times 7.6 = 91.2 \ cm^2$

Now,

Area is also given by $QN \times PS$

$\Rightarrow 91.2 = QN \times 8$

$\Rightarrow QN = \frac{91.2}{8} = 11.4 \ cm$

Therefore, value of QN is  $11.4 \ cm$

We know that

Area of parallelogram  $= base \times height$

Here,

Base of parallelogram(AB) = 35 cm

and

Height of parallelogram(DL) = h cm

$\Rightarrow 1470 = 35 \times h$

$\Rightarrow h = \frac{1470}{35} = 42 \ cm$

Similarly,

Area is also given by $AD \times BM$

$\Rightarrow 1470 = 49 \times BM$

$\Rightarrow BM = \frac{1470}{49} = 30 \ cm$

Therefore, the value of BM and DL are  is  30cm and 42cm respectively

$\inline \Delta ABC$. Also find the length of$\inline AD$.

We know that

Area of triangle $= \frac{1}{2} \times base \times height$

Now,

When base = 5 cm and height = 12 cm

Then, the area is equal to

$\Rightarrow \frac{1}{2} \times 5 \times 12 = 30 \ cm^2$

Now,

When base = 13 cm and height = AD area remain same

Therefore,

$\Rightarrow 30= \frac{1}{2} \times 13 \times AD$

$\Rightarrow AD = \frac{60}{13} \ cm$

Therefore, value of AD is  $\frac{60}{3} \ cm$  and the area is equal to $30 \ cm^2$

$\inline \Delta ABC.$  What will be the height from $C$ to $AB$ i.e., $CE$ ?

We know that

Area of triangle $= \frac{1}{2} \times base \times height$

Now,

When base(BC) = 9 cm and height(AD) = 6 cm

Then, the area is equal to

$\Rightarrow \frac{1}{2} \times 9 \times 6 = 27 \ cm^2$

Now,

When base(AB) = 7.5 cm and height(CE) = h , area remain same

Therefore,

$\Rightarrow 27= \frac{1}{2} \times 7.5 \times CE$

$\Rightarrow CE = \frac{54}{7.5}= 7.2 \ cm$

Therefore, value of CE is  7.2cm  and area is equal to $27 \ cm^2$

Question:(a)  In Fig Which square has a larger perimeter?

The outer square has a larger perimeter. Since each side of the inner square forms a triangle. The length of the third side of a triangle is less than the sum of the other two lengths.

The arc length of the circle is slightly greater than the side length of the inner square. Therefore circumference of the inner circle is greater than the inner square

NCERT solutions for class 7 maths chapter 11 perimeter and area-Exercise: 11.3

$14 \; cm$

We know that

Circumference  of a circle is $= 2\pi r$

$\Rightarrow 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \ cm$

Therefore, the circumference of the circle is  88 cm

$28\; mm$

We know that

Circumference  of circle is $= 2\pi r$

$\Rightarrow 2\pi r = 2 \times \frac{22}{7} \times 28 = 176 \ mm$

Therefore, the circumference of the circle is  176 mm

$21\; cm$

We know that

Circumference of circle is $= 2\pi r$

$\Rightarrow 2\pi r = 2 \times \frac{22}{7} \times 21 = 132 \ cm$

Therefore, circumference of circle is  132 cm

Question:

$radius=14\; mm$  (Take $\pi =\frac{22}{7}$)

We know that

Area of circle is $= \pi r^2$

$\Rightarrow \pi r^2 = \frac{22}{7}\times (14)^2 = 616 \ mm^2$

Therefore, the area of the circle is  $616 \ mm^2$

Question:

$diameter=49\; m$

We know that

Area of circle is $= \pi r^2$

$\Rightarrow \pi r^2 = \frac{22}{7}\times \left ( \frac{49}{2} \right )^2 =1886.5 \ m^2$

Therefore, area of circle is  $1886.5 \ m^2$

Question:

$radius=5\; cm$

We know that

Area of circle is $= \pi r^2$

$\Rightarrow \pi r^2 = \frac{22}{7}\times \left (5 \right )^2 =\frac{550}{7} \ cm^2$

Therefore, the area of the circle is  $\frac{550}{7} \ cm^2$

It is given that  circumference of a circular sheet is  154 m

We know that

Circumference of circle is $= 2\pi r$

$\Rightarrow 154 = 2\pi r$

$\Rightarrow 154 = 2 \times \frac{22}{7} \times r$

$\Rightarrow r= \frac{49}{2}= 24.5 \ m$

Now,

Area of circle $= \pi r^2$

$\Rightarrow \frac{22}{7}\times \left ( 24.5 \right )^2 = 1886.5 \ m^2$

Therefore, the radius and area of the circle are 24.5 m and $1886.5 \ m^2$  respectively

It is given that diameter of a circular garden is $21 \; m$.

We know that

Circumference of circle is $= 2\pi r$

$\Rightarrow 2 \times \frac{22}{7}\times \frac{21}{2} = 66 \ m$

Now, length of the rope requires to makes $2$ rounds of fence is

$\Rightarrow 2 \times$  circumference of circle

$\Rightarrow 2 \times 66 = 132 \ m$

Now,  cost of rope at $Rs.4 \; per \; meter$ is

$\Rightarrow 132 \times 4 = 528 \ Rs$

Therefore, length of the rope requires to makes $2$ rounds of fence is  132 m and   cost of rope at $Rs.4 \; per \; meter$ is  Rs 528

We know that

Area of circle $= \pi r^2$

Area of circular sheet with radius 4 cm $= \3.14 \times (4)^2 = 50.24 \ cm^2$

Area of the circular sheet with radius 3 cm $= \3.14 \times (3)^2 = 28.26 \ cm^2$

Now,

Area of remaining sheet = Area of circle with radius 4 cm - Area od circle with radius 3 cm

$= 50.24 - 28.26 = 21.98\ cm^2$

Therefore, Area of remaining sheet is  $21.98\ cm^2$

It is given that diameter of a circular table is 1.5m.

We know that

Circumference of circle is $= 2\pi r$

$\Rightarrow 2 \times \frac{22}{7}\times \frac{1.5}{2} = 4.71 \ m$

Now, length of the lace required is

$\Rightarrow$  circumference of circle $= 4.71 \ m$

Now,  cost of lace at $Rs.15 \; per \; meter$ is

$\Rightarrow 4.71 \times 15 = 70.65 \ Rs$

Therefore, length of the lace required is  4.71 m and   cost of lace at  $Rs.15 \; per \; meter$  is  Rs 70.65

It is given that the diameter of semi-circle is 10 cm.

We know that

Circumference of semi circle is $= \pi r$

Circumference of semi-circle with diameter 10 cm  including diameter is

$\Rightarrow \left ( \frac{22}{7}\times \frac{10}{2} \right )+10 = 15.7+10=25.7 \ cm$

Therefore, Circumference of semi-circle with diameter 10 cm  including diameter is 25.7 cm

It is given that the diameter of a circular table is 1.6m.

We know that

Area of circle is $= \pi r^2$

$\Rightarrow 3.14 \times \left ( \frac{1.6}{2} \right )^2 = 2.0096 \ m^2$

Now, the cost of polishing at $Rs.15 \; per \; m^2$ is

$\Rightarrow 2.0096 \times 15 = 30.144 \ Rs$

Therefore, the cost of polishing at  $Rs.15 \; per \; m^2$   is  Rs 30.144

It is given that the length of wire is 44 cm

Now, we know that

Circumference of the circle (C) = $2 \pi r$

$\Rightarrow 44 = 2 \pi r$

$\Rightarrow r = \frac{44\times 7}{2 \times 22} = 7 \ cm$

Now,

Area of circle (A) = $\pi r^2$

$\Rightarrow \pi r^2= \frac{22}{7}\times (7)^2= 154 \ cm^2$                                      - (i)

Now,

Perimeter of square(P) = $4a$

$\Rightarrow 44=4a$

$\Rightarrow a = \frac{44}{4} = 11 \ cm$

Area of sqaure = $a^2$

$\Rightarrow a^2 = (11)^2= 121 \ cm^2$                                                   -(ii)

From equation (i) and (ii) we can clearly see that area of the circular-shaped wire is more than square-shaped wire

It is given that radius of circular card sheet is $14\; cm$

Now, we know that

Area of circle (A) = $\pi r^2$

$\Rightarrow \pi r^2= \frac{22}{7}\times (14)^2= 616 \ cm^2$                                      - (i)

Now,

Area of circle with radius 3.5 cm is

$\Rightarrow \pi r^2= \frac{22}{7}\times (3.5)^2= 38.5 \ cm^2$

Area of two such circle is = $38.5 \times 2 = 77 \ cm^2$                  -(ii)

Now, Area of rectangle = $l \times b$

$\Rightarrow l \times b = 3 \times 1 = 3\ cm^2$                                                   -(iii)

Now, the remaining area is (i) - [(ii) + (iii)]

$\Rightarrow 616- [77+3]\Rightarrow 616-80 = 536 \ cm^2$

Therefore,  area of the remaining sheet  is  $536 \ cm^2$

(Take  $\pi =3.14$)

It is given that the radius of the circle is $2 \ cm$

Now, we know that

Area of the circle (A) = $\pi r^2$

$\Rightarrow \pi r^2= 3.14 \times(2)^2= 12.56 \ cm^2$                                      - (i)

Now,

Now, Area of square  = $a^2$

$\Rightarrow a^2 = (6)^2 = 36 \ cm^2$                                                   -(ii)

Now, the remaining area is (ii) - (i)

$\Rightarrow 36 - 12.56 = 23.44 \ cm^2$

Therefore, the area of the remaining aluminium sheet  is  $23.44 \ cm^2$

It is given that circumference of circle is $31.4 \ cm$

Now, we know that

Circumference of circle is = $2 \pi r$

$\Rightarrow 31.4 = 2 \pi r$

$\Rightarrow 31.4 = 2 \times 3.14 \times r$

$\Rightarrow r = 5 \ cm$

Now, Area of circle (A) = $\pi r^2$

$\Rightarrow \pi r^2= 3.14 \times(5)^2= 78.5 \ cm^2$

Therefore, radius and  area of the circle  are $5 \ cm \ and \ 78.5 \ cm^2$  respectively

It is given that the diameter of the flower bed is $66\; m$

Therefore, $r = \frac{66}{2} = 33 \ m$

Now, we know that

Area of the circle (A) = $\pi r^2$

$\Rightarrow \pi r^2= 3.14 \times(33)^2= 3419.46\ m^2$                             -(i)

Now, Area of outer circle with radius(r ') = 33 + 4 = 37 cm is

$\Rightarrow \pi r'^2= 3.14 \times(37)^2= 4298.66 \ m^2$                               -(ii)

Area of the path is equation (ii) - (i)

$\Rightarrow 4298.66-3419.46 = 879.2 \ m^2$

Therefore, the area of the path is  $879.2 \ m^2$

It is given that the radius of the sprinkler is $12 m$

Now, we know that

Area of the circle (A) = $\pi r^2$

Area cover by sprinkle is

$\Rightarrow \pi r^2= 3.14 \times(12)^2= 452.16 \ m^2$

And the area of the flower garden is $314\; m^{2}$

Therefore,  YES  sprinkler water the entire garden

We know that

Circumference of circle = $2 \pi r$

Now, the circumference of the inner circle  with radius (r) = $19-10=9 \ m$ is

$\Rightarrow 2 \pi r = 2 \times 3.14 \times 9 = 56.52 \ m$

And the circumference of the outer circle  with radius (r ') = 19 m is

$\Rightarrow 2 \pi r = 2 \times 3.14 \times 19 = 119.32 \ m$

Therefore, the circumference of inner and outer circles are $56.52 \ m \ and \ 119.32 \ m$  respectively

It is given that radius of wheel is $28\; cm$

Now, we know that

Circumference of circle = $2 \pi r$

$\Rightarrow 2 \pi r = 2 \times 3.14 \times 28 = 175.84 \ cm$

Now, number of rotation done by wheel to go 352 m  is

$\Rightarrow \frac{352 \ m}{175.84 \ cm} = \frac{35200}{175.84}\cong 200$

Therefore,  number of rotation done by wheel to go 352 m is  200

It is given that minute hand of a circular clock is $15\; cm$ long  i.e. ( r = 15 cm)

Now, we know that one hour means a complete circle of minute hand

Now,

Circumference of circle = $2 \pi r$

$\Rightarrow 2 \pi r = 2 \times 3.14 \times 15 = 94.2 \ cm$

Therefore,  distance cover by minute hand in one hour is  94.2 cm

Solutions of NCERT for class 7 maths chapter 11 perimeter and area topic 11.6 conversion of units

Question:(i) Convert the following:

$50\; cm^{2} \; in\; mm^{2}$

1 cm = 10 mm

$1 cm^2=1cm\times1cm=10mm\times 10 mm=100mm^2$

therefor

$50 cm^2=50\times 100mm^2=5000mm^2$

Question:(ii)

$2\; ha\; in\; m^{2}$

ha represents hectare

1 ha is 10000 m2

therefor

$2ha=2\times10000m^2=20000m^2$

Question:(iii)

$10\; m^{2}\; in\; cm^{2}$

1m = 100cm

$1m^2=1m\times1m=100cm\times100cm=10000cm^2$

Therefor

$10m^2=10\times10000cm^2=100,000cm^2$

Question:(iv) Convert the following:

$1000\; cm^{2}\; in\; m^{2}$

The conversion is done as follows

$\\100 cm = 1m\\ 100\times100cm^2=1\times1 m^2\\10000cm^2=1m^2\\1000cm^2=0.1m^2$

CBSE NCERT solutions for class 7 maths chapter 11 perimeter and area-Exercise: 11.4

Answer: It is given that the garden is $90\; m$ long and $75\; m$ broad

It is clear that it is rectangular shaped with $lenght = 90 \ m \ and \ breath = 75 \ m$

We know that the area of the rectangle is $= lenght \times breath$

$\Rightarrow lenght \times breath = 90 \times 75 = 6750 \ m^2 = 0.675 \ ha$                   -(i)

Now, length and breadth of the outer rectangle is

Area of the outer rectangle is

$\Rightarrow lenght \times breath = 100 \times 85 = 8500 \ m^2$                              -(ii)

Area of the path is (ii) - (i)

$\Rightarrow 8500 - 6750 = 1750 \ m^2$

Therefore, the area of the path is $1750 \ m^2$

It is given that park is  $125\; m$ long and  $65\; m$ broad

It is clear that it is rectangular shaped with $length = 125 \ m \ and \ breadth = 65 \ m$

We know that area of rectangle is $= length \times breadth$

$\Rightarrow length \times breadth = 125 \times 65 = 8125 \ m^2$                   -(i)

Now, length and breadth of outer rectangle is

Area of outer rectangle is

$\Rightarrow length \times breadth = 131 \times 71 = 9301 \ m^2$                              -(ii)

Area of path is (ii) - (i)

$\Rightarrow 9301 - 8125 = 1176 \ m^2$

Therefore,  area of path is $1176 \ m^2$

It is given that cardboard is $8\; cm$ long and $5\; cm$ wide such that there is a margin of $1.5\; cm$ along each of its sides

It is clear that it is rectangular shaped with $length = 8 \ cm \ and \ breadth = 5 \ cm$

We know that the area of the rectangle is $= length \times breadth$

$\Rightarrow length \times breadth = 8 \times 5 = 40 \ cm^2$                   -(i)

Now, the length and breadth of cardboard without margin is

Area of cardboard without margin is

$\Rightarrow length \times breadth = 5 \times 2 = 10 \ cm^2$                              -(ii)

Area of margin is (i) - (ii)

$\Rightarrow 40-10= 30 \ cm^2$

Therefore, the area of margin is  $30 \ cm^2$

It is given that room  is $5.5\; m$ long and $4\; m$ wide.

It is clear that it is rectangular shaped with  $length = 5.5 \ m \ and \ breadth = 4 \ m$

We know that the area of the rectangle is $= length \times breadth$

$\Rightarrow length \times breadth = 5.5 \times 4 = 22 \ m^2$                   -(i)

Now, when verandah of width $2.25\; m$  is constructed all along outside the room then length and breadth of the room  is

Area of the room after verandah of width $2.25\; m$  is constructed

$\Rightarrow length \times breadth = 10 \times 8.5 = 85 \ m^2$                              -(ii)

Area of verandah is (ii) - (i)

$\Rightarrow 85-22 = 63 \ m^2$

Therefore, the area of the verandah is  $63 \ m^2$

It is given that room  is $5.5\; m$ long and $4\; m$ wide.

It is clear that it is rectangular shaped with  $length = 5.5 \ m \ and \ breadth = 4 \ m$

We know that the area of the rectangle is $= length \times breadth$

$\Rightarrow length \times breadth = 5.5 \times 4 = 22 \ m^2$                   -(i)

Now, when verandah of width $2.25\; m$  is constructed all along outside the room then length and breadth of the room  is

Area of the room after verandah of width $2.25\; m$  is constructed

$\Rightarrow length \times breadth = 10 \times 8.5 = 85 \ m^2$                              -(ii)

Area of verandah is (ii) - (i)

$\Rightarrow 85-22 = 63 \ m^2$

Therefore,  area of verandah is  $63 \ m^2$

Now, the cost of cementing the floor of the verandah at the rate of $Rs.200 \; per\; m^{2}$ is

$\Rightarrow 63 \times 200 = 12600 \ Rs$

Therefore,  cost of cementing the floor of the verandah at the rate of $Rs.200 \; per\; m^{2}$ is $Rs \ 12600$

Is is given that side of square garden is $30\; m$

We know that area of square is = $a^2$

$\Rightarrow a^2 =(30)^2 =900 \ m^2$                   -(i)

Now, area of square garden without 1 m boarder is

$\Rightarrow a^2 =(28)^2 =784 \ m^2$                              -(ii)

Area of path is (i) - (ii)

$\Rightarrow 900-784 = 116 \ m^2$

Therefore,  area of path is  $116 \ m^2$

Is is given that side of square garden is $30\; m$

We know that area of square is = $a^2$

$\Rightarrow a^2 =(30)^2 =900 \ m^2$                   -(i)

Now, area of square garden without 1 m boarder is

$\Rightarrow a^2 =(28)^2 =784 \ m^2$                              -(ii)

Area of path is (i) - (ii)

$\Rightarrow 900-784 = 116 \ m^2$

Therefore,  area of path is  $116 \ m^2$

Now,  cost of planting grass in the remaining portion of the garden at the rate of $Rs. 40\; per\; m^{2}$ is

$\Rightarrow 784 \times 40 = 31360 \ Rs$

Therefore,  cost of planting grass in the remaining portion of the garden at the rate of $Rs. 40\; per\; m^{2}.$  is  $Rs \ 31360$

It is given that width of each road is $10m$ and the length of rectangular park is $700\; m$ and breadth is $300\; m$

Now, We know that area of rectangle is = $length \times breadth$

Area of total park is

$\Rightarrow 700 \times 300 = 210000 \ m^2$                   -(i)

Area of road parallell to width of the park ( ABCD ) is

$\Rightarrow 300 \times 10 = 3000 \ m^2$                           -(ii)

Area of road parallel to length of park ( PQRS )  is

$\Rightarrow 700 \times 10 = 7000 \ m^2$                          -(iii)

The common area of both the roads ( KMLN ) is

$\Rightarrow 10 \times 10 = 100 \ m^2$                               -(iv)

Area of roads = $[(ii)+(iii)-(iv)]$

$\Rightarrow 7000+3000-100=9900 \ m^2 = 0.99 \ ha$                -(v)

Now, Area of the park excluding crossroads is = $[(i)-(v)]$

$\Rightarrow 210000-9900 = 200100 \ m^2 = 20.01 \ ha$

It is given that the width of each road is $3m$ and length of rectangular park is $90\; m$ and breadth is $60\; m$

Now, We know that area of rectangle is = $length \times breadth$

Area of total park is

$\Rightarrow 90 \times 60 = 5400 \ m^2$                   -(i)

Area of road parallell to width of the park ( ABCD ) is

$\Rightarrow 60 \times 3 = 180 \ m^2$                           -(ii)

Area of road parallel to length of park ( PQRS )  is

$\Rightarrow 3 \times 90 = 270 \ m^2$                          -(iii)

The common area of both the roads ( KMLN ) is

$\Rightarrow 3 \times 3 = 9 \ m^2$                               -(iv)

Area of roads = $[(ii)+(iii)-(iv)]$

$\Rightarrow 180+270-9 = 441 \ m^2$

Therefore, the area of road is  $441 \ m^2$

It is given that the width of each road is $3m$ and the length of rectangular park is $90\; m$ and breadth is $60\; m$

Now, We know that area of rectangle is = $length \times breadth$

Area of the total park is

$\Rightarrow 90 \times 60 = 5400 \ m^2$                   -(i)

Area of road parallel to the width of the park ( ABCD ) is

$\Rightarrow 60 \times 3 = 180 \ m^2$                           -(ii)

Area of road parallel to the length of park ( PQRS )  is

$\Rightarrow 3 \times 90 = 270 \ m^2$                          -(iii)

The common area of both the roads ( KMLN ) is

$\Rightarrow 3 \times 3 = 9 \ m^2$                               -(iv)

Area of roads = $[(ii)+(iii)-(iv)]$

$\Rightarrow 180+270-9 = 441 \ m^2$

Now, the cost of constructing the roads at the rate of  $Rs. 110\; per\; m^{2}$ is

$\Rightarrow 441 \times 110 = 48510 \ Rs$

Therefore, the cost of constructing the roads at the rate of  $Rs. 110\; per\; m^{2}$ is $Rs \ 48510$

It is given that the radius of the circle is 4 cm

We know that circumference of circle = $2 \pi r$

$\Rightarrow 2 \pi r = 2 \times 3.14 \times 4 = 25.12 \ cm$

And perimeter of the square is $= 4a$

$\Rightarrow 4a = 4 \times 4 = 16 \ cm$

Length of cord left is $= 25.12 - 16 = 9.12 \ cm$

Therefore, Length of cord left is  $9.12 \ cm$

We know that area rectangle is = $length \times breadth$

Area of rectangular land with length 10 m and width 5 m is

$\Rightarrow length \times breadth 10 \times 5 = 50 \ m^2$

Therefore, area of rectangular land with length 10 m and width 5 m is $50 \ m^2$

We know that area of circle is = $\pi r^2$

Area of flower bed with radius 2 m is

$\Rightarrow \pi r^2 = 3.14 \times (2)^2 = 12.56 \ m^2$

Therefore, area of flower bed with radius 2 m is $12.56 \ m^2$

Area of the lawn excluding the area of the flower bed = Area of rectangular lawn - Area of flower bed

= $50 - 12.56 = 37.44 \ m^2$

Therefore, area of the lawn excluding the area of the flower bed is $37.44 \ m^2$

We know that circumference of the circle is = $2 \pi r$

Circumference of the flower bed with radius 2 m is

$\Rightarrow 2 \pi r = 2 \times 3.14 \times 2 = 12.56 \ m$

Therefore, the circumference of the flower bed with radius 2 m is  $12.56 \ m$

Area  of shaded portion = Area of the whole rectangle ( ABCD ) - Area of two triangles ( AFE and BCE)

Area of the rectangle with length 18 cm and width 10 cm is

$\Rightarrow length \times breadth = 18 \times 10 = 180 \ cm^2$

Area of triangle AFE with base 10 cm and height 6 cm is

$\Rightarrow \frac{1}{2} \times \ base \times height =\frac{1}{2} \times 10 \times 6 = 30 \ cm^2$

Area of triangle BCE with base 8 cm and height 10 cm is

$\Rightarrow \frac{1}{2} \times \ base \times height =\frac{1}{2} \times 8 \times 10 = 40 \ cm^2$

Now, Area  of the shaded portion is

$\Rightarrow 180 - (40+30)$

$\Rightarrow 180 - 70 = 110 \ cm^2$

Question:

Area  of shaded portion = Area of the whole square ( PQRS ) - Area of three triangles ( PTQ , STU and QUR )

Area of the square with side 20cm is

$\Rightarrow a^2 = (20)^2 =400 \ cm^2$

Area of triangle PTQ with base 10 cm and height 20 cm is

$\Rightarrow \frac{1}{2} \times \ base \times height =\frac{1}{2} \times 10 \times 20 = 100 \ cm^2$

Area of triangle STU with base 10 cm and height 10 cm is

$\Rightarrow \frac{1}{2} \times \ base \times height =\frac{1}{2} \times 10 \times 10 = 50 \ cm^2$

Area of triangle QUR with base 20 cm and height 10 cm is

$\Rightarrow \frac{1}{2} \times \ base \times height =\frac{1}{2} \times 20 \times 10 = 100 \ cm^2$

Now, Area  of the shaded portion is

$\Rightarrow 400 - (100+100+50)$

$\Rightarrow 400-250=150 \ cm^2$

Area  of  quadrilateral ABCD = Area of triangle ABC + Area of tringle ADC

Area of triangle ABC with base 22 cm and height 3 cm is

$\Rightarrow \frac{1}{2} \times \ base \times height =\frac{1}{2} \times 22 \times 3 = 33 \ cm^2$

Area of triangle ADX with base 22 cm and height 3 cm is

$\Rightarrow \frac{1}{2} \times \ base \times height =\frac{1}{2} \times 22 \times 3 = 33 \ cm^2$

Therefore, the area of quadrilateral ABCD is = $33+33=66 \ cm^2$

NCERT Solutions for Class 7 Maths- Chapter-wise

 Chapter No. Chapter Name Chapter 1 Solutions of NCERT for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 Solutions of NCERT for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry

NCERT Solutions for Class 7- Subject-wise

Some important point from NCERT solutions for class 7 maths chapter 11 perimeter and area which you should remember-

Area of a triangle: If the base length and height of a triangle are given

$Area=\frac{1}{2}\times base\times height$

Perimeter of the triangle:- Perimeter will be equal to the sum the sides of the triangle

$\text{Perimeter}=a+b+c\\$

a- First side of the triangle

b- Second side of the triangle

c- Third side of the triangle

Area of Circles:-

$Area=\pi r^2$

Circumference of Circle:-

$\text{Circumference }=2\pi r$

Area of a parallelogram:

$Area= b\times h$

b-> base length

h-> height

Tip- You shouldn't just memorize the formulas, also understand the concept of how these formulas are derived. If you go through solutions of NCERT for class 7 maths chapter 11 perimeter and area, you will understand all these concepts very easily.