NCERT solutions for class 7 maths chapter 11(Perimeter and Area), in this chapter
we discuss about following topics. You can direct access from the given link:
- Solution Exercise 11.1
- Solution Exercise 11.2
- Solution Exercise 11.3
- Solution Exercise 11.4
- Download the Solution in pdf of Exercise 11.1, Exercise 11.2, Exercise 11.3 and Exercise 11.4
7th Maths Exercise 11.1 Solution
Direct access ncert solutions for class 7 maths chapter 11 ,the student can download the solution in pdf from the link given above. There for they can understand the concept easily. Click here for solution Exercise 11.2 ,
Question 1:
The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find:
(i) its area (ii) the cost of the land, if 1 m2 of the land costs Rs. 10,000.
Answer 1:
(i) It is known that the area of a rectangle is = (length×breadth)
We have given length = 500 m and Breadth = 300 m
Therefore area = (500×300) = (150000) =150000 m2
(ii) Cost of 1 m2 land is Rs.10000
So cost of rectangular field = 150000×10000=1,500,000,000
Hence area of rectagular field is 150000 m2 and cost is Rs. 1,500,000,000
Question 2:
Find the area of a square park whose perimeter is 320 m.
Answer 2:
It is known that perimeter of square = 4×side
Therefore 4×side = 320 m
Now Area of square = (side)2
= (80)2 = 6400 m2
Hence area of park is 6400 m2
Question 3:
Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.
Answer 3:
We have given area = 440 m2 and length = 22 m
It is known that area of rectangle = (length×breadth)
= (22×breadth)= 440 m2
breadth= 440/22=20 m
Now perimeter of rectangle =2(length+breadth)
=2(22m+20m)=2(42m) = 84m
Hence breadth of rectangle is 20m and Perimeter is 84m.
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 4:
We have given perimeter = 100cm and length = 35m
Perimeter of rectangle =2(length+breadth)
=2(35m+breadth) = 100m
(35m+breadth) =100/2=50m
breadth = 50m – 35m = 15m
It is known that area of rectangle =(length×breadth)
=(35×15)=525 m2
Hence Breadth is 15m and area is 525 m2
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 5:
We have given side of square = 60m …..(i)
and length of rectangle = 90m ……(ii)
Area of square = (side)2 and Area of rectangle =(length×breadth)
According to question,
Area of square = Area of rectangle
(side)2 =(length×breadth)
(60)2 = (90 ×breadth)
3600 m2 =(90m×breadth)
breadth= 3600/90 = 40m
Hence Breadth of rectangular park 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 6:
We have given length of rectangle = 40cm and breadth = 22cm
If wire is in shape of rectangle, then length of wire will be equal to perimeter of rectangle,
Therefore
Perimeter of rectangle =2(length+breadth)
=2(40+22)=2(62)=124cm
So length of wire is 124cm
Now it is rebent in square form so,
Perimeter of square = 124cm
4×side = 124cm
side= 124/4=31cm
Now area of Square = (side)2 = (31)2 = 961cm2
Area of rectangle =length × breadth
= 40×22=880 cm2
Hence side of square is 31cm and square inclosed 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 7:
We have given Perimeter of rectangle = 130cm and breadth = 30cm
Perimeter of rectangle =2(length+breadth)
130=2(length+30cm)
(length+30cm) = 130/2 = 65cm
length = 65-30 = 35cm
Now Area of rectangle =length ×breadth=35×30=1050 cm2
Hence length of rectangle is 35cm and area of rectangle is 1050 cm2
Question 8:
A door of length 2 m and breadth 1m 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 m2.
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 m2.
Answer 8:
We have given length of door = 2m and breadth of door = 1m
And length of wall= 4.5m and breadth of wall = 3.6m
Now, Area of wall =length ×breadth= 4.5×3.6 = 16.2 m2
Ans area of door = length ×breadth = 2×1=2 m2
Area for painting (white wash) = Area of wall – Area of door
=16.2-2=14.2 m2
This is required area for white washing.
Now cost of white wash =14.2×20=284
Hence the cost of white washing the wall is Rs. 284
7th Maths Exercise 11.2 Solution
In ncert solutions for class 7 maths chapter 11 ,the student can download the solution in pdf from the link given above . There for they can understand the concept easily. Click here for solution Exercise 11.1
Question 1:
Find the area of each of the following parallelograms:
Answer 1:
(a) Area of parallelogram = (Base ×Height)
=7×4=28 cm2
(b) Area of parallelogram = (Base ×Height)
=5×3 =15 cm2
(c) Area of parallelogram = (Base ×Height)
=3.5×2.5=8.75 cm2
(d) Area of parallelogram = (Base ×Height)
=5×4.8=24 cm2
(e) Area of parallelogram =(Base ×Height)
=2×4.4=8.8 cm2
Question 2:
Find the area of each of the following triangles:
Answer 2:
Question 3:
| S. No | Base | Height | Area of the Parallelogram |
| a. | 20cm | ______ | 246 cm2 |
| b. | _____ | 15 cm | 154.5 cm2 |
| c. | _____ | 8.4 cm | 48.72 cm2 |
| d. | 15.6 cm | ______ | 16.38 cm2 |
Answer 3:
| S. No | Base | Height | Area of the Parallelogram |
| a. | 20cm | 12.3 cm | 246 cm2 |
| b. | 10.3cm | 15 cm | 154.5 cm2 |
| c. | 5.8 cm | 8.4 cm | 48.72 cm2 |
| d. | 15.6 cm | 1.05 cm | 16.38 cm2 |
Question 4:
Find the missing values:
| Base | Height | Area of Triangle |
| 15 cm | _____ | 87 cm2 |
| ____ | 31.4 mm | 1256 mm2 |
| 22 cm | ____ | 170.5 cm2 |
Answer 4:
| Base | Height | Area of Triangle |
| 15 cm | 11.6 cm | 87 cm2 |
| 80 mm | 31.4 mm | 1256 mm2 |
| 22 cm | 15.5 cm | 170.5 cm2 |
Question 5:
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:
(a) the area of the parallegram PQRS (b) QN, if PS = 8 cm
Answer 5:
(a) We have given, SR (base) = 12 cm and QM (height) = 7.6 cm
Area of parallelogram = (Base ×Height)
= (12 ×7.6)= 91.2cm2
(b) We have given, PS = 8 cm
If we take PS as base then, QN will be height therefore area of parallelogram
=(8×Height) = 91.2 cm2
Height =91.2/8=11.4 cm
Hence QN = 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 cm2 , AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Answer 6:
If we take AD as base then BM will be height, therefore
Area of parallelogram = 1470 cm2
⇒ (Base ×Height) = 1470
⇒ 49×MB = 1470
⇒ MB=1470/49 =30 cm
We have given AB = 35 cm, AD = 49 cm and area of parallelogram = 1470 cm2
If we take AB as base then DL will be height, therefore
Area of parallelogram = 1470 cm2
⇒ (Base ×Height) = 1470
⇒ 35×DL = 1470
⇒ DL=1470/35= 42 cm
Hence length of MB = 30 cm and DL = 42 cm
Question 7:
∆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 ∆ABC. Also find the length of AD.
Answer 7:
If we take AB (5cm) as base then AC (12m) will be height, Therefore
Area of triangle =1/2(Base × Height)
=1/2 (5 ×12) =30 cm2
Now if we take BC (13cm) as base then AD will be height, therefore
Area of triangle = 30 cm2
⇒1/2(Base ×Height) = 30 cm2
⇒ 1/2 (13×AD)=30 cm2
⇒ AD=(30×2)/13 = 60/13 cm
Hence Area of triangle ABC = 30 cm2 and AD = 60/13 cm
Question 8:
∆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 ∆ABC. What will be the height from C to AB i.e., CE?
Answer 8:
Now if we take AB (7.5 cm ) as base then EC will be height of the triangle, Therefore
Area of triangle = 27 cm2
Hence area of triangle = 27 cm2 and height of the triangle from C is 7.2 cm
Some definition of the chapter NCERT solutions for class 7 maths chapter 11, which used in solving the question.
Perimeter is the distance around a closed figure whereas area is the part of plane
occupied by the closed figure.
We have learnt how to find perimeter and area of a square and rectangle in the earlier class. They are:
(a) Perimeter of a square = 4 × side
(b) Perimeter of a rectangle = 2 × (length + breadth)
(c) Area of a square = side × side
(d) Area of a rectangle = length × breadth