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.
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
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: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
Question:1(i) The length and the breadth of a rectangular piece of land are and respectively. Find its area
Answer:
It is given that the length and the breadth of a rectangular piece of land are and
Now, we know that
Area of the rectangle (A) =
Therefore, area of rectangular piece of land is
Question:1(ii) The length and the breadth of a rectangular piece of land are and respectively. Find the cost of the land, if of the land costs
Answer:
It is given that the length and the breadth of a rectangular piece of land are and
Now, we know that
Area of rectangle (A) =
Now, it is given that of the land costs
Therefore, cost of of land is
Question:2 Find the area of a square park whose perimeter is .
Answer:
It is given that the perimeter of the square park is
Now, we know that
The perimeter of a square is (P) , where a is the side of the square
Now,
Area of the square (A)
Therefore, the area of a square park is
Answer:
It is given that the area of rectangular land is and the length is
Now, we know that
Area of rectangle is
Now,
The perimeter of the rectangle is
Therefore, breath and perimeter of the rectangle are 20m and 88m respectively
Question:4 The perimeter of a rectangular sheet is . If the length is , find its breadth. Also find the area.
Answer:
It is given that perimeter of a rectangular sheet is and length is
Now, we know that
The perimeter of the rectangle is
Now,
Area of rectangle is
Therefore, breath and area of the rectangle are 15cm and respectively
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 and the length of the rectangular park is
Now, we know that
Area of square
Area of rectangle
Area of square = Area of rectangle
Therefore, breadth of the rectangle is 40 m
Answer:
It is given that the length of rectangular wire is and breadth is .
Now, if it reshaped into a square wire
Then,
The perimeter of rectangle = perimeter of the square
Now,
Area of rectangle
Area of square
Therefore, the side of the square is 31 cm and we can clearly see that squareshaped wire encloses more area
Answer:
It is given that the perimeter of a rectangle is and breadth is
Now, we know that
The perimeter of the rectangle is
Now,
Area of rectangle is
Therefore, the length and area of the rectangle are 35cm and respectively
Answer:
It is given that the length of door is and breadth is and the length of the wall is and the breadth is
Now, we know that
Area of rectangle is
Thus, the area of the wall is
And
Area of the door is
Now, Area to be painted = Area of wall  Area of door = 16.2  2 = 14.2
Now, the cost of whitewashing the wall, at the rate of is
Therefore, the cost of whitewashing the wall, at the rate of 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
Answer:
The total area of rectangle
i) The first rectangle is divided into 6 equal parts. So the area of each part will be onesixth of the total area
ii) The rectangle is divided into 4 equal parts, area of each part= oneforth 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 oneeighth of the total area of rectangle = 3 cm^{2}
Question:1(i) Find the area of following parallelograms:
Answer:
Area of the parallelogram is the product of base and height
Question:1(ii) Find the area of following parallelograms:
Answer:
Area of the parallelogram is the product of base and height
Question:(iii) Find the area of the following parallelograms:
In a parallelogram , and the perpendicular from on is
Answer:
Area of parallelogram = the product of base and height
Question:1(a) Find the area of the following parallelograms:
Answer:
We know that
Area of parallelogram
Here,
Base of parallelogram = 7cm
and
Height of parallelogram = 4 cm
Therefore, the area of the parallelogram is
Question:1(b) Find the area of the following parallelograms:
Answer:
We know that
Area of parallelogram
Here,
Base of parallelogram = 5cm
and
Height of parallelogram = 3 cm
Therefore, the area of the parallelogram is
Question:1(c) Find the area of the following parallelograms:
Answer:
We know that
Area of parallelogram
Here,
Base of parallelogram = 2.5cm
and
Height of parallelogram = 3.5 cm
Therefore, area of parallelogram is
Question:1(d) Find the area of the following parallelograms:
Answer:
We know that
Area of parallelogram
Here,
Base of parallelogram = 5cm
and
Height of parallelogram = 4.8 cm
Therefore, the area of a parallelogram is
Question:1(e) Find the area of the following parallelograms:
Answer:
We know that
Area of parallelogram
Here,
Base of parallelogram = 2 cm
and
Height of parallelogram = 4.4 cm
Therefore, area of parallelogram is
Question:2(a) Find the area of each of the following triangles:
Answer:
We know that
Area of triangle
Here,
Base of triangle = 4 cm
and
Height of triangle =3 cm
Therefore, area of triangle is
Question:2(b) Find the area of the following triangles:
Answer:
We know that
Area of triangle
Here,
Base of triangle = 5 cm
and
Height of triangle =3.2 cm
Therefore, the area of the triangle is
Question:2(c) Find the area of the following triangles:
Answer:
We know that
Area of triangle
Here,
Base of triangle = 3 cm
and
Height of triangle =4 cm
Therefore, area of triangle is
Question:2(d) Find the area of the following triangles:
Answer:
We know that
Area of triangle
Here,
Base of triangle = 3 cm
and
Height of triangle =2 cm
Therefore, the area of the triangle is
Question:3 Find the missing values:
Answer:
We know that
Area of parallelogram
a) Here, base and area of parallelogram is given
b) Here height and area of parallelogram is given
c) Here height and area of parallelogram is given
d) Here base and area of parallelogram is given
Question:4 Find the missing values:
Answer:
We know that
Area of triangle
a) Here, the base and area of the triangle is given
b) Here height and area of the triangle is given
c) Here base and area of the triangle is given
Question:5(a) is a parallelogram (Fig 11.23). is the height from to and is the height from to . If and
Answer:
We know that
Area of parallelogram
Here,
Base of parallelogram = 12 cm
and
Height of parallelogram = 7.6 cm
Therefore, area of parallelogram is
Question:5(b) is a parallelogram (Fig 11.23). is the height from to and is the height from to . If and
Answer:
We know that
Area of parallelogram
Here,
Base of parallelogram = 12 cm
and
Height of parallelogram = 7.6 cm
Now,
Area is also given by
Therefore, value of QN is
Question:6 and are the heights on sides and respectively, of parallelogram (Fig 11.24). If the area of the parallelogram is and find the length of and
Answer:
We know that
Area of parallelogram
Here,
Base of parallelogram(AB) = 35 cm
and
Height of parallelogram(DL) = h cm
Similarly,
Area is also given by
Therefore, the value of BM and DL are is 30cm and 42cm respectively
Question:7 is right angled at (Fig 11.25). is perpendicular to . If and Find the area of
Answer:
We know that
Area of triangle
Now,
When base = 5 cm and height = 12 cm
Then, the area is equal to
Now,
When base = 13 cm and height = AD area remain same
Therefore,
Therefore, value of AD is and the area is equal to
Question:8 is isosceles with and (Fig 11.26). The height from to is Find the area of
What will be the height from to i.e., ?
Answer:
We know that
Area of triangle
Now,
When base(BC) = 9 cm and height(AD) = 6 cm
Then, the area is equal to
Now,
When base(AB) = 7.5 cm and height(CE) = h , area remain same
Therefore,
Therefore, value of CE is 7.2cm and area is equal to
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
Question:1(a) Find the circumference of the circles with the following radius: (Take )
Answer:
We know that
Circumference of a circle is
Therefore, the circumference of the circle is 88 cm
Question:1(b) Find the circumference of the circles with the following radius: (Take )
Answer:
We know that
Circumference of circle is
Therefore, the circumference of the circle is 176 mm
Question:1(c) Find the circumference of the circles with the following radius: (Take )
Answer:
We know that
Circumference of circle is
Therefore, circumference of circle is 132 cm
Question:2(a) Find the area of the following circles, given that:
Answer:
We know that
Area of circle is
Therefore, the area of the circle is
Question:2(b) Find the area of the following circles, given that:
Answer:
We know that
Area of circle is
Therefore, area of circle is
Question:2(c) Find the area of the following circles, given that:
Answer:
We know that
Area of circle is
Therefore, the area of the circle is
Answer:
It is given that circumference of a circular sheet is 154 m
We know that
Circumference of circle is
Now,
Area of circle
Therefore, the radius and area of the circle are 24.5 m and respectively
Answer:
It is given that diameter of a circular garden is .
We know that
Circumference of circle is
Now, length of the rope requires to makes rounds of fence is
circumference of circle
Now, cost of rope at is
Therefore, length of the rope requires to makes rounds of fence is 132 m and cost of rope at is Rs 528
Answer:
We know that
Area of circle
Area of circular sheet with radius 4 cm
Area of the circular sheet with radius 3 cm
Now,
Area of remaining sheet = Area of circle with radius 4 cm  Area od circle with radius 3 cm
Therefore, Area of remaining sheet is
Answer:
It is given that diameter of a circular table is 1.5m.
We know that
Circumference of circle is
Now, length of the lace required is
circumference of circle
Now, cost of lace at is
Therefore, length of the lace required is 4.71 m and cost of lace at 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 semicircle is 10 cm.
We know that
Circumference of semi circle is
Circumference of semicircle with diameter 10 cm including diameter is
Therefore, Circumference of semicircle with diameter 10 cm including diameter is 25.7 cm
Question:8 Find the cost of polishing a circular tabletop of diameter , if the rate of polishing is . (Take )
Answer:
It is given that the diameter of a circular table is 1.6m.
We know that
Area of circle is
Now, the cost of polishing at is
Therefore, the cost of polishing at is Rs 30.144
Answer:
It is given that the length of wire is 44 cm
Now, we know that
Circumference of the circle (C) =
Now,
Area of circle (A) =
 (i)
Now,
Perimeter of square(P) =
Area of sqaure =
(ii)
From equation (i) and (ii) we can clearly see that area of the circularshaped wire is more than squareshaped wire
Answer:
It is given that radius of circular card sheet is
Now, we know that
Area of circle (A) =
 (i)
Now,
Area of circle with radius 3.5 cm is
Area of two such circle is = (ii)
Now, Area of rectangle =
(iii)
Now, the remaining area is (i)  [(ii) + (iii)]
Therefore, area of the remaining sheet is
Answer:
It is given that the radius of the circle is
Now, we know that
Area of the circle (A) =
 (i)
Now,
Now, Area of square =
(ii)
Now, the remaining area is (ii)  (i)
Therefore, the area of the remaining aluminium sheet is
Question:12 The circumference of a circle is . Find the radius and the area of the circle? (Take )
Answer:
It is given that circumference of circle is
Now, we know that
Circumference of circle is =
Now, Area of circle (A) =
Therefore, radius and area of the circle are respectively
Answer:
It is given that the diameter of the flower bed is
Therefore,
Now, we know that
Area of the circle (A) =
(i)
Now, Area of outer circle with radius(r ') = 33 + 4 = 37 cm is
(ii)
Area of the path is equation (ii)  (i)
Therefore, the area of the path is
Answer:
It is given that the radius of the sprinkler is
Now, we know that
Area of the circle (A) =
Area cover by sprinkle is
And the area of the flower garden is
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 )
Answer:
We know that
Circumference of circle =
Now, the circumference of the inner circle with radius (r) = is
And the circumference of the outer circle with radius (r ') = 19 m is
Therefore, the circumference of inner and outer circles are respectively
Question:16 How many times a wheel of radius must rotate to go ? (Take )
Answer:
It is given that radius of wheel is
Now, we know that
Circumference of circle =
Now, number of rotation done by wheel to go 352 m is
Therefore, number of rotation done by wheel to go 352 m is 200
Question:17 The minute hand of a circular clock is long. How far does the tip of the minute hand move in hour. (Take)
Answer:
It is given that minute hand of a circular clock is long i.e. ( r = 15 cm)
Now, we know that one hour means a complete circle of minute hand
Now,
Circumference of circle =
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:
Answer:
1 cm = 10 mm
therefor
Question:(ii) Convert the following:
Answer: It is given that the garden is long and broad
It is clear that it is rectangular shaped with
We know that the area of the rectangle is
(i)
Now, length and breadth of the outer rectangle is
Area of the outer rectangle is
(ii)
Area of the path is (ii)  (i)
Therefore, the area of the path is
Answer:
It is given that park is long and broad
It is clear that it is rectangular shaped with
We know that area of rectangle is
(i)
Now, length and breadth of outer rectangle is
Area of outer rectangle is
(ii)
Area of path is (ii)  (i)
Therefore, area of path is
Answer:
It is given that cardboard is long and wide such that there is a margin of along each of its sides
It is clear that it is rectangular shaped with
We know that the area of the rectangle is
(i)
Now, the length and breadth of cardboard without margin is
Area of cardboard without margin is
(ii)
Area of margin is (i)  (ii)
Therefore, the area of margin is
Question:4(i) A verandah of width is constructed all along outside a room which is long and wide. Find: the area of the verandah.
Answer:
It is given that room is long and wide.
It is clear that it is rectangular shaped with
We know that the area of the rectangle is
(i)
Now, when verandah of width is constructed all along outside the room then length and breadth of the room is
Area of the room after verandah of width is constructed
(ii)
Area of verandah is (ii)  (i)
Therefore, the area of the verandah is
Question:4(ii) A verandah of width is constructed all along outside a room which is long and wide. Find: the cost of cementing the floor of the verandah at the rate of .
Answer:
It is given that room is long and wide.
It is clear that it is rectangular shaped with
We know that the area of the rectangle is
(i)
Now, when verandah of width is constructed all along outside the room then length and breadth of the room is
Area of the room after verandah of width is constructed
(ii)
Area of verandah is (ii)  (i)
Therefore, area of verandah is
Now, the cost of cementing the floor of the verandah at the rate of is
Therefore, cost of cementing the floor of the verandah at the rate of is
Question:5(i) A path wide is built along the border and inside a square garden of side . Find: the area of the path.
Answer:
Is is given that side of square garden is
We know that area of square is =
(i)
Now, area of square garden without 1 m boarder is
(ii)
Area of path is (i)  (ii)
Therefore, area of path is
Question:5(ii) A path wide is built along the border and inside a square garden of side. Find: the cost of planting grass in the remaining portion of the garden at the rate of
Answer:
Is is given that side of square garden is
We know that area of square is =
(i)
Now, area of square garden without 1 m boarder is
(ii)
Area of path is (i)  (ii)
Therefore, area of path is
Now, cost of planting grass in the remaining portion of the garden at the rate of is
Therefore, cost of planting grass in the remaining portion of the garden at the rate of is
Answer:
It is given that width of each road is and the length of rectangular park is and breadth is
Now, We know that area of rectangle is =
Area of total park is
(i)
Area of road parallell to width of the park ( ABCD ) is
(ii)
Area of road parallel to length of park ( PQRS ) is
(iii)
The common area of both the roads ( KMLN ) is
(iv)
Area of roads =
(v)
Now, Area of the park excluding crossroads is =
Question:7(i) Through a rectangular field of length and breadth , 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 , find the area covered by the roads.
Answer:
It is given that the width of each road is and length of rectangular park is and breadth is
Now, We know that area of rectangle is =
Area of total park is
(i)
Area of road parallell to width of the park ( ABCD ) is
(ii)
Area of road parallel to length of park ( PQRS ) is
(iii)
The common area of both the roads ( KMLN ) is
(iv)
Area of roads =
Therefore, the area of road is
Question:7(ii) Through a rectangular field of length and breadth, 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 .find the cost of constructing the roads at the rate of .
Answer:
It is given that the width of each road is and the length of rectangular park is and breadth is
Now, We know that area of rectangle is =
Area of the total park is
(i)
Area of road parallel to the width of the park ( ABCD ) is
(ii)
Area of road parallel to the length of park ( PQRS ) is
(iii)
The common area of both the roads ( KMLN ) is
(iv)
Area of roads =
Now, the cost of constructing the roads at the rate of is
Therefore, the cost of constructing the roads at the rate of is
Answer:
It is given that the radius of the circle is 4 cm
We know that circumference of circle =
And perimeter of the square is
Length of cord left is
Therefore, Length of cord left is
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 =
Area of rectangular land with length 10 m and width 5 m is
Therefore, area of rectangular land with length 10 m and width 5 m is
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 =
Area of flower bed with radius 2 m is
Therefore, area of flower bed with radius 2 m is
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
=
Therefore, area of the lawn excluding the area of the flower bed is
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 =
Circumference of the flower bed with radius 2 m is
Therefore, the circumference of the flower bed with radius 2 m is
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
Area of triangle AFE with base 10 cm and height 6 cm is
Area of triangle BCE with base 8 cm and height 10 cm is
Now, Area of the shaded portion is
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
Area of triangle PTQ with base 10 cm and height 20 cm is
Area of triangle STU with base 10 cm and height 10 cm is
Area of triangle QUR with base 20 cm and height 10 cm is
Now, Area of the shaded portion is
Question:11 Find the area of the quadrilateral . Here, and
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
Area of triangle ADX with base 22 cm and height 3 cm is
Therefore, the area of quadrilateral ABCD is =
NCERT Solutions for Class 7 Maths Chapterwise
Chapter No. 
Chapter Name 
Chapter 1 

Chapter 2 
CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals 
Chapter 3 

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 
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
Perimeter of the triangle: Perimeter will be equal to the sum the sides of the triangle
a First side of the triangle
b Second side of the triangle
c Third side of the triangle
Area of Circles:
r> Radius of the circle
Circumference of Circle:
r> Radius of the circle
Area of a parallelogram:
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!!!