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

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.

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:

454541

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:

1643019243331

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:

1643019339998

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:

1414

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:

4344

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

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

Therefore, area of parallelogram is 91.2 \ cm^2

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.

454546

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

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:

23233

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

NCERT Solutions for Class 7 Subject Wise

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. 

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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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