NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area

 

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

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

11.1 Introduction

11.2 Squares And Rectangles

11.2.1 Triangles As Parts Of Rectangles

11.2.2 Generalising For Other Congruent Parts Of Rectangles

11.3 Area Of A Parallelogram

11.4 Area Of A Triangle

11.5 Circles

11.5.1 Circumference Of A Circle

11.5.2 Area Of Circle

11.6 Conversion Of Units

11.7 Applications

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area is given below:

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

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

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 100cm2 and perimeter is 50 cm. A rectangle with sides 50 cm and 2 cm ha area = 100cm2  but perimeter is 104 cm

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

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 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, 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 is3.6\; 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}

            

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 is3.6\; m 

Now, we know that

Area of rectangle is = length \times breath

Thus, the area of the wall is

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

And

Area of the door is

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

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

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

\Rightarrow 14.2 \times 20 = 284 \ Rs

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

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

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.

            

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 cm2

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

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

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

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

  

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

CBSE 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:

             

Answer:

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:

            

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:

            

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:

            

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, area of parallelogram is  8.8 \ cm^2

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

            

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:

            

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:

            

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:

            

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:

             

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:

            

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. IfSR=12 \; cm and

             QM=7.6 \; cm. Find:

              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. IfSR=12 \; cm and

             QM=7.6 \; cm. Find:

              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,

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.

            

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 ofAD.

          

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, 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 fromA to BC, is 6\; cm, Find the area of

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

            

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

Solutions of NCERT for class 7 maths chapter 11 perimeter and area topic 11.5.1 circumference of a circle

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

                 

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?

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} ).

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 costsRs.15. (Take \pi =3.14)

Answer:

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

We know that

Circumference of circle is = 2\pi r

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

Now, length of the lace required is 

\Rightarrow  circumference of circle = 4.71 \ m

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

\Rightarrow 4.71 \times 15 = 70.65 \ Rs

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

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

             

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} )

         

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)

               

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)

              

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

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

Question:(i) Convert the following:

 50\; cm^{2} \; in\; mm^{2}

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 m2

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

CBSE NCERT solutions for class 7 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



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



Area of outer rectangle is 

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

Area of path is (ii) - (i)

\Rightarrow 9301 - 8125 = 1176 \ m^2

Therefore,  area of path is 1176 \ m^2

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


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


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


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)



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 side30\; 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 is given that side of square garden is 30\; m

We know that area of square is = a^2

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



Now, area of square garden without 1 m boarder is

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

Area of path is (i) - (ii)

\Rightarrow 900-784 = 116 \ m^2

Therefore,  area of path is  116 \ m^2

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

\Rightarrow 784 \times 40 = 31360 \ Rs

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

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:



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.

Answer:

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\; mand breadth60\; 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}.

Answer:



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: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)

             

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.

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.

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.

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.

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:

               

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

               

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

NCERT Solutions for Class 7 Maths- Chapter-wise 

Chapter No.

Chapter Name

Chapter 1

Solutions of NCERT for class 7 maths chapter 1 Integers

Chapter 2

CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals

Chapter 3

NCERT solutions for class 7 maths chapter 3 Data Handling

Chapter 4

Solutions of NCERT for class 7 maths chapter 4 Simple Equations

Chapter 5

CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles

Chapter 6

NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties

Chapter 7

Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles

Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities

Chapter 9

CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers

Chapter 10

NCERT solutions for class 7 maths chapter 10 Practical Geometry

Chapter 11

Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area

Chapter 12

CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions

Chapter 13

NCERT solutions for class 7 maths chapter 13 Exponents and Powers

Chapter 14

Solutions of NCERT for class 7 maths chapter 14 Symmetry

 NCERT Solutions for Class 7- Subject-wise 

Solutions of NCERT for class 7 maths

CBSE NCERT solutions for class 7 science

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

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

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

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

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

a- First side of the triangle

b- Second side of the triangle

c- Third side of the triangle

Area of Circles:-  

                      Area=\pi r^2

 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 for class 7 maths chapter 11 perimeter and area, you will understand all these concepts very easily.

Happy Reading!!! 

 

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