NCERT Solutions for Class 7 Maths Chapter 11
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.1, Exercise 11.2, Exercise 11.3 and Exercise 11.4 in English Medium or Prashnavali 11.1, Prashnavali 11.2
2, Prashnavali 11.3 and Prashnavali 11.4 in Hindi Medium to study online or download in PDF file format. Latest NCERT Books for 2020-2021 and NCERT Solutions Offline Apps for the new academic session 2020-2021 are also available to free download without any registration.NCERT Solutions for Class 7 Maths Chapter 11
| Class: 7 | Maths (English and Hindi Medium) |
| Chapter 11: | Perimeter and Area |
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7 Maths Chapter 11 Solutions
Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.1, Exercise 11.2, Exercise 11.3 and Exercise 11.4 with step by step proper answers and solutions are given below. NCERT Solutions 2020-21 are updated for the new session 2020-21 based on new NCERT Books.
7 Maths Chapter 11 All Exercises in English Medium
7 Maths Chapter 11 All Exercises in Hindi Medium
Class 7 Maths Exercise 11.1 and Exercise 11.2 Solution in Video
Class 7 Maths Exercise 11.3 and Exercise 11.4 Solution in Video
About NCERT Solutions for Class 7 Maths Chapter 11
In 7 Maths Chapter 11 Perimeter and Area, we have to go through calculation of perimeters of some basic figures like triangular region, square and some other simple figures. Area of parallelogram, area of triangle and area of triangle as a part of a quadrilateral, etc. We must consider the fact that all the congruent triangles are equal in area but the triangles equal in area need not be congruent. As we know that the distance around a circular region is known as its circumference, so the perimeter of a circular region is normally refer as circumference. Important formulae related to areas are given below:
1. Perimeter of a square = 4 × side
2. Perimeter of a rectangle = 2 × (length + breadth)
3. Area of a square = side × side
4. Area of a rectangle = length × breadth
5. Area of a parallelogram = base × height
6. Area of a triangle = ½ (area of the parallelogram generated from it) = ½ × base × height
7. Circumference of a circle = πd, where d is the diameter of a circle
8. Area of a circle = πr², where r is the radius of the circle.
The value of π can be taken 22/7 or 3.14 (approximately). Use any value if the value of π is not mention in the question.
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Important Questions on Class 7 Maths Chapter 11
Breadth of a rectangular piece of land = 300 m
Area of a rectangular piece of land
= Length x Breadth
= 500 x 300
= 1,50,000 m^2
पथ सहित बगीचे की लंबाई = 90 + 5 + 5 = 100 m
पथ सहित बगीचे की चौड़ाई = 75 + 5 + 5 = 85 m
पथ सहित बगीचे का क्षेत्रफल
= लंबाई x चौड़ाई
= 100 x 85
= 8,500 m^2
केवल बगीचे का क्षेत्रफल
= लंबाई x चौड़ाई
= 90 x 75
= 6,750 m^2
इसलिए, पथ का क्षेत्रफल
= पथ सहित बगीचे का क्षेत्रफल – केवल बगीचे का क्षेत्रफल
= 8,500 – 6,750
= 1,750 m^2
4 x side = 320
side = 320/4 = 80 m
Now,
Area of square park
= side x side
= 80 x 80
= 6400 m^2
Thus, the area of square park is 6400 m^2.
length x breadth = 440 m^2
22 x breadth = 440
breadth = 440/22 = 20 m
Now,
Perimeter of rectangular park
= 2 (length + breadth)
= 2 (22 + 20)
= 2 x 42 = 84 m
Thus, the perimeter of rectangular park is 84 m.
आयताकार पार्क की चौड़ाई = 65 m और पथ की चौड़ाई = 3 m
पथ सहित पार्क की लंबाई = 125 + 3 + 3 = 131 m
पथ सहित पार्क की चौड़ाई = 65 + 3 + 3 = 71 m
पथ का क्षेत्रफल
= पथ सहित पार्क का क्षेत्रफल – केवल पार्क का क्षेत्रफल
= (131 x 71) – (125 x 65)
= 9301 – 8125
= 1,176 m^2
अतः, पथ का क्षेत्रफल 1,176 m^2 है।
इसकी प्रत्येक भुजाओं के अनुदिश 1.5 cm चौड़ा हाशिया छोड़ा गया है।
पेंटिग की लंबाई में कमी = 8 – (1.5 + 1.5) = 8 – 3 = 5 cm
और पेंटिग की चौड़ाई में कमी = 5 – (1.5 + 1.5) = 5 – 3 = 2 cm
हाशिये का कुल क्षेत्रफल
= गत्ते (ABCD) का क्षेत्रफल – गत्ते (EFGH) का क्षेत्रफल
= (8 x 5) – (5 x 2)
= 40 – 10
= 30 cm^2
अतः, हाशिये का कुल क्षेत्रफल 30 cm^2 है।
2 (length + breadth) = 100 cm
2 (35 + breadth) = 100
35 + breadth = 100/2
35 + breadth = 50
breadth = 50 – 35
breadth = 15 cm
Now,
Area of rectangular sheet
= length x breadth
= 35 x 15 = 525 cm^2
Thus, breadth and area of rectangular sheet are 15 cm and 525 cm^2 respectively.
2 (length + breadth) = 130 cm
2 (length + 30) = 130
length + 30 = 130/2
length + 30 = 65
length = 65 – 30 = 35 cm
Now area of rectangle
= length x breadth
= 35 x 30
= 1050 cm^2
Thus, the area of rectangle is 1050 cm^2.
= 3.14 x 12 x 12
= 3.14 x 144
= 452.16 m^2
वृत्ताकार फूलों के बगीचे का क्षेत्रफल = 314 m2
यहाँ, वृत्ताकार फूलों के बगीचे का क्षेत्रफल, फव्वारे द्वारा छिड़काव किए गए क्षेत्रफल से कम है।
अतः, फव्वारा पूरे बगीचे में पानी का छिड़काव कर सकेगा।
वृत्ताकार घड़ी की परिधि
= 2πr
= 2 x 3.14 x 15
= 94.2 cm
अतः, मिनट की सुई की नोक 1 घंटे में 94.2 cm दूरी तय करती है।
