NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area - Download PDF

Edited By Ravindra Pindel | Updated on Feb 07, 2024 06:10 PM IST

NCERT Solutions for Perimeter and area class 7 - 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. 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.

Latest: NCERT Class 7 Maths: Chapterwise Important Formulas with Examples and Points

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NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7 - Important Formulae
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area - Important Points
NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7
NCERT Solution for Class 7 Maths Chapter 11 Perimeter and Area (Intext Questions and Exercise)
NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7 Topic 11.2
NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7 Exercise: 11.1
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.2.2
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.3
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.5.1
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.6
NCERT Solutions for Class 7th Maths Chapter 11 Perimeter and Area Exercise 11.4
Perimeter And Area Class 7 Maths Chapter 11-Topics

You should try to solve all the NCERT problems including examples for a better understanding of the concepts. In CBSE NCERT solution 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 Classes 6 to 12 by clicking on the above link. Here you will get solutions to four exercises of this chapter. Students are supposed to refer to the NCERT Class 7 Syllabus and know the exam pattern and important topics.

NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7 - Important Formulae

Perimeter of a Triangle = Side₁ + Side₂ + Side₃

Perimeter of a Square = 4 × side of square

Perimeter of a Rectangle = 2 × ( Length + Breadth or Width )

Circumference or Perimeter of a Circle = πd = 2πr

Where d = Diameter of circle = 2r, r = Radius of circle

Area of a Square = Side × Side

Area of rectangle = Length × Breadth

Congruent parts of rectangle:

The area of each congruent part = (1/2) × ( The area of the rectangle )

Area of a parallelogram = Base × Height

Area of a Triangle = (1/2) × ( Base) × ( Height)

All the congruent triangles are equal in the area but the triangles equal in the area need not be congruent.

Area of a circle = πr2 where, r = Radius of circle

Unit Conversions:

1 cm² = 100 mm²

1 m² = 10000 cm²

1 hectare = 10000 m2

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area - Important Points

Perimeter: The total length of the boundary of a two-dimensional shape.

Circumference: The distance around the boundary of a circle.

Area: The measure of the space enclosed by a two-dimensional shape.

Congruent: Two shapes are congruent if they have the same size and shape.

Base: The bottom side of a two-dimensional shape, like a parallelogram or triangle.

Height: The perpendicular distance between the base and the opposite side in a two-dimensional shape.

Diameter: The longest chord (line segment connecting two points on a circle) passing through the center of a circle.

Radius: The distance from the center of a circle to any point on its boundary.

Unit Conversions: Converting measurements from one unit to another, based on their relationships.

Free download NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area PDF for CBSE Exam.

NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7

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NCERT Solution for Class 7 Maths Chapter 11 Perimeter and Area (Intext Questions and Exercise)

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

Question:1 What would you need to find, area or perimeter, to answer the following?

How much space does a blackboard occupy?

Answer: The space of the board includes the whole area of the board

Question:2 What would you need to find, area or perimeter, to answer the following?

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

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

Question:3 What would you need to find, area or perimeter, to answer the following?

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

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

Question:4 What would you need to find, area or perimeter, to answer the following?

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

Answer: Plastic sheet need to cover is the area of the rectangular swimming pool

Question:2 Give two examples where the area increases as the perimeter increases.

Answer: A square of side 1m has perimeter 4 m and area 1 m ² . When all sides are increased by 1m then perimeter =8m and area= 4 m ² . 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

Question:3 Give two examples where the area does not increase when perimeter increases.

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

NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7 Exercise: 11.1

Question: 1(i) The length and the breadth of a rectangular piece of land are $500\; m$ and $300\; m$ respectively. Find its area

Answer: 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 $150000 \ m ^2$

Question: 1(ii) The length and the breadth of a rectangular piece of land are $500\; m$ and $300\; m$ respectively. Find the cost of the land, if $1\; m^{2}$ of the land costs $Rs.10,000$

Answer: 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) = $l \times b = 500\times 300 = 150000 \ m ^2$

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

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

Question: 2 Find the area of a square park whose perimeter is $320\; m$ .

Answer: 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 $6400 \ m^2$

Question: 3 Find the breadth of a rectangular plot of land, if its area is $440\; m^{2}$ and the length is $22\; m$ . Also find its perimeter.

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

Question: 4 The perimeter of a rectangular sheet is $100\; cm$ . If the length is $35\; cm$ , find its breadth. Also find the area.

Answer: 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

Question: 5 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

Answer: 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

Question: 6 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?

Answer: 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

Question: 7 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.

Answer: 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

Question: 8 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 $3.6\; m$ (Fig11.6). Find the cost of white washing the wall, if the rate of white washing the wall is $Rs.20\; per\; m^{2}$

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Answer: 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

Question 1: Each of the following rectangles of length 6 cm and breadth 4 cm is composed of congruent polygons. Find the area of each polygon.

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Answer:

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 cm ²

iii) and (iv) are divided into two equal parts. Area of each part will be one half the total area of rectangle= 12 cm ²

v) Area of rectangle is divided into 8 equal parts. Area of one part is one-eighth of the total area of rectangle = 3 cm ²

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

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

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Answer: Area of the parallelogram is the product of base and height

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

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

Answer: Area of the parallelogram is the product of base and height

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

Question:(iii) Find the area of the following parallelograms:

In a parallelogram $ABCD$ , $AB=7.2\; cm$ and the perpendicular from $C$ on $AB$ is $4.5\; cm$

Answer: Area of parallelogram = the product of base and height

$=7.2\times4.5=32.4cm^2$

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

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

Answer: 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: 1(b) Find the area of the following parallelograms:

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Answer: 1643019312227

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: 1(c) Find the area of the following parallelograms:

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Answer: 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: 1(d) Find the area of the following parallelograms:

45454

Answer: 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: 1(e) Find the area of the following parallelograms:

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Answer: 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, the area of a parallelogram is $8.8 \ cm^2$

Question: 2(a) Find the area of each of the following triangles:

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Answer: 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: 2(b) Find the area of the following triangles:

36363

Answer: 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: 2(c) Find the area of the following triangles:

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Answer: 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: 2(d) Find the area of the following triangles:

225455

Answer: 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: 3 Find the missing values:

454141

Answer: 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: 4 Find the missing values:

4544654

Answer: 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

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

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

Question: 5(a) $PQRS$ is a parallelogram (Fig 11.23). $QM$ 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:

454548 the area of the parallelogram $PQRS$

Question: 5(b) $PQRS$ is a parallelogram (Fig 11.23). $QM$ 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:

75757575 $QN$ , if $PS=8\; cm$ .

Answer: 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,

The 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 $11.4 \ cm$

Question: 6 $DL$ and $BM$ are the heights on sides $AB$ and $AD$ respectively, of parallelogram $ABCD$ (Fig 11.24). If the area of the parallelogram is $1470\; cm^{2}.$ $AB=35\; cm$ and $AD=49\; cm,$ find the length of $BM$ and $DL.$

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Answer: 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

Question:7 $\Delta ABC$ is right angled at $A$ (Fig 11.25). $AD$ is perpendicular to $BC$ . If $AB=5\; cm,$ $BC=13\; cm$ and $AC=12\; cm,$ Find the area of

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

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

Question: 8 $\Delta ABC$ is isosceles with $AB=AC=7.5\; cm$ and $BC=9\; cm$ (Fig 11.26). The height $AD$ from $A$ to $BC,$ is $6\; cm,$ Find the area of

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

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Answer: 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 $27 \ cm^2$

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.5.1

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

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Answer: 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.

Question:(b) In Fig 11.31, Which is larger, perimeter of smaller square or the circumference of the circle?

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Answer: 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

Question: 1(a) Find the circumference of the circles with the following radius: (Take $\pi =\frac{22}{7}$ )

$14 \; cm$

Answer: 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

Question: 1(b) Find the circumference of the circles with the following radius: (Take $\pi =\frac{22}{7}$ )

$28\; mm$

Answer: 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

Question: 1(c) Find the circumference of the circles with the following radius: (Take $\pi =\frac{22}{7}$ )

$21\; cm$

Answer: 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: 2(a) Find the area of the following circles, given that:

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

Answer: 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: 2(b) Find the area of the following circles, given that:

$diameter=49\; m$

Answer: 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: 2(c) Find the area of the following circles, given that:

$radius=5\; cm$

Answer: 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 $\frac{550}{7} \ cm^2$

Question: 3 If the circumference of a circular sheet is $154\; m,$ find its radius. Also, find the area of the sheet. (Take $\pi =\frac{22}{7}$ )

Answer: 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

Question: 4 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 cost of the rope, if it costs $Rs.4 \; per \; meter.$ (Take $\pi =\frac{22}{7}$ ).

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Answer: 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

Question: 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 $\pi =3.14$ )

Answer: 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 $21.98\ cm^2$

Question: 6 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 $Rs.15.$ (Take $\pi =3.14$ )

Answer: It is given that the 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

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

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Answer: 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

Question: 8 Find the cost of polishing a circular table-top of diameter $1.6\; m$ , if the rate of polishing is $Rs.\; 15/m^{2}$ . (Take $\pi =3.14$ )

Answer: 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

Question: 9 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 area, the circle or the square? (Take $\pi =\frac{22}{7}$ )

Answer: 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

Question: 10. 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 adjoining figure). Find the area of the remaining sheet. (Take $\pi =\frac{22}{7}$ )

23232

Answer: 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 $536 \ cm^2$

Question:11 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 $\pi =3.14$ )

Answer: 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 $23.44 \ cm^2$

Question: 12 The circumference of a circle is $31.4\; cm$ . Find the radius and the area of the circle? (Take $\pi =3.14$ )

Answer: 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

Question: 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? $(\pi =3.14)$

255

Answer: 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 $879.2 \ m^2$

Question:14 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? (Take $\pi =3.14$ )

Answer: 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

Question:15 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take $\pi =3.14$ )

1215

Answer: 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

Question:16 How many times a wheel of radius $28\; cm$ must rotate to go $352\; m$ ? (Take $\pi =\frac{22}{7}$ )

Answer: 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

Question: 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 $\pi =3.14$ )

Answer: 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

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.6

Question:(i) Convert the following:

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

Answer: 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) Convert the following:

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

Answer: ha represents hectare

1 ha is 10000 m ²

therefor

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

Question:(iii) Convert the following:

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

Answer: 1m = 100cm

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

Therefor

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

Question:(iv) Convert the following:

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

Answer: The conversion is done as follows

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

NCERT Solutions for Class 7th Maths Chapter 11 Perimeter and Area Exercise 11.4

Question: 1 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.

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

1643019827233
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 $1750 \ m^2$

Question: 2 A $3\; m$ wide path runs outside and around a rectangular park of length $125\; m$ and breadth $65\; m$ . Find the area of the path.

Answer: 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

1643019878729
Area of the 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 $1176 \ m^2$

Question: 3 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

Answer: 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

1643019936958 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 $30 \ cm^2$

Question: 4(i) A verandah of width $2.25\; m$ is constructed all along outside a room which is $5.5\; m$ long and $4\; m$ wide. Find: the area of the verandah

Answer: 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

1643019978746 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 $63 \ m^2$

Question: 4(ii) A verandah of width $2.25\; m$ is constructed all along outside a room which is $5.5\; m$ long and $4\; m$ wide. Find: the cost of cementing the floor of the verandah at the rate of $Rs.200 \; per\; m^{2}$ .

Answer: 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

1658484613627 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 $Rs \ 12600$

Question: 5(i) A path $1\; m$ wide is built along the border and inside a square garden of side $30\; m$ . Find: the area of the path.

Answer: 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)

1643020143340
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 $116 \ m^2$

Question: 5(ii) A path $1\; m$ wide is built along the border and inside a square garden of side $30\; m$ . Find: the cost of planting grass in the remaining portion of the garden at the rate of $Rs. 40\; per\; m^{2}.$

Answer: 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)

1658484696990
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 $Rs \ 31360$

Question: 6 Two cross roads, each of width $10m$ , 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.

Answer:

1643020535544
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 = $[(i)-(v)]$

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

Question: 7(i) 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 the area covered by the roads.

download2525

Answer:

1658484791265

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 $441 \ m^2$

Question: 7(ii) 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 the cost of constructing the roads at the rate of $Rs. 110\; per\; m^{2}$ .

download2525

Answer:

1658484870635
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 $Rs \ 48510$

Question: 8 Pragya wrapped a cord around a circular pipe of the 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 cord left? $(\pi =3.14)$

12122121

Answer: 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 $9.12 \ cm$

Question: 9(i) The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: the area of the whole land.

112121

Answer: 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 $50 \ m^2$

Question: 9(ii) The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: the area of the flower bed.

112121

Answer: 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 $12.56 \ m^2$

Question: 9(iii) The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: the area of the lawn excluding the area of the flower bed.

112121

Answer: 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 $37.44 \ m^2$

Question: 9(iv) The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: the circumference of the flower bed

112121

Answer: 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 $12.56 \ m$

Question: 10(i) In the following figure, find the area of the shaded portion:

Answer: 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: 10(ii) In the following figure, find the area of the shaded portion:

2121212

Answer: 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

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

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

Question: 11 Find the area of the quadrilateral $ABCD$ . Here, $AC=22\; cm,$ $BM=3\; cm,$ $DN=3\; cm,$ and $BM\perp AC,DN\perp AC$

123115

Answer: 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 = $33+33=66 \ cm^2$

Perimeter And Area Class 7 Maths Chapter 11-Topics

Squares And Rectangles
Triangles As Parts Of Rectangles
Generalising For Other Congruent Parts Of Rectangles
Area Of A Parallelogram
Area Of A Triangle
Circles
Circumference Of A Circle
Area Of Circle
Conversion Of Units
Applications

NCERT Solutions for Class 7 Maths Chapter Wise

Chapter No.	Chapter Name
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

NCERT Solutions for Class 7 Subject Wise

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Science

Some important point from NCERT solutions for Class 7th Maths chapter 11 Perimeter and Area

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

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

The 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:-

$Area=\pi r^2$

r-> Radius of the circle

Circumference of Circle:-

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

r-> Radius of the circle

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 Solutions for Class 7 , you will understand all these concepts very easily.

Happy Reading!!!

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. Is NCERT maths chapter perimeter and area important

Yes the NCERT chapter perimeter and area is important. The concepts studied in this chapter will be used in the coming classes. Therefore you should practice ncert solution for class 7 maths chapter 11. these solutions will help you to get deeper understanding of the concepts. Also you can download perimeter and area class 7 pdf.

2. What is area?

The part of the plane or region occupied by a closed figure. The concepts related to area are discussed in chapter 11 maths class 7. Students can download class 7th maths chapter 11 pdf and study both offline and online. you can study class 7 hindi chapter 11 pdf also.

3. What are the topics covered in NCERT solution Maths Class 7 Chapter11?

Here is the topics that are covered in NCERT solution Maths Class 7 Chapter11

Squares And Rectangles
Triangles As Parts Of Rectangles
Generalising For Other Congruent Parts Of Rectangles
Area Of A Parallelogram
Area Of A Triangle
Circles
Circumference Of A Circle
Area Of Circle
Conversion Of Units
Applications

Also Read

NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story

NCERT Solutions for Class 7 Science Chapter 17 Forests Our Lifeline

NCERT Solutions for Class 7 Science Chapter 16 Water A Precious Resource

NCERT Solutions for Class 7 Science Chapter 14 Electric Current and Its Effects

NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

NCERT Solutions for Class 7 Science Chapter 12 Reproduction in Plants

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area - Download PDF

NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7 - Important Formulae

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area - Important Points

NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7

NCERT Solution for Class 7 Maths Chapter 11 Perimeter and Area (Intext Questions and Exercise)

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

NCERT Solutions for Maths Chapter 11 Perimeter and Area Class 7 Exercise: 11.1

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.2.2

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

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

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.5.1

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

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.6

NCERT Solutions for Class 7th Maths Chapter 11 Perimeter and Area Exercise 11.4

Perimeter And Area Class 7 Maths Chapter 11-Topics

NCERT Solutions for Class 7 Maths Chapter Wise

NCERT Solutions for Class 7 Subject Wise

Some important point from NCERT solutions for Class 7th Maths chapter 11 Perimeter and Area