NCERT Solutions For Class 7 Maths Chapter 11

NCERT Solutions For Class 7 Maths Chapter 11 PDF Free Download

NCERT Solutions for Class 7 Maths Chapter 11, Perimeter and Area, are provided here so that students can check for the solutions for each question, whenever they are facing difficulty while solving from NCERT class 7 maths book. These solutions for chapter 11, are available in PDF format so that students can download it and learn offline as well and also they can use it as a worksheet for practice.

The NCERT solutions for Class 7 contains the different chapters with the various solutions to the problems. Students have to practice mathematics on a regular basis and become an expert in the subject. Students of Class 7 are suggested to solve perimeter and area class 7 questions, in order to strengthen the fundamentals and be able to solve questions that are usually asked in the final exams. Among all chapters in grade 7 maths subject, Perimeter and Area is an important topic for the students, whose concepts are used in further studies. One needs to practice thoroughly these topics from chapter 11, to score well in the examination.

Class 7 Maths NCERT Solutions – Perimeter and Area

NCERT Solutions Class 7 Maths Perimeter and Area, PDF is provided here are in a detailed manner, where one can find a step-by-step solution to all types of questions which will be asked in the main exams. These PDF solutions are prepared by our subject experts in accordance with CBSE syllabus (2018-2019) and NCERT curriculum, to assist students in their exam preparations.

The topics covered in chapter 11 of 7th class Maths are;

  • Perimeter and Area of Squares and Rectangles
  • Area of Triangle
  • Area of Parallelogram
  • Triangles as Parts of Rectangles
  • Generalising for other Congruent Parts of Rectangles
  • Circumference of a Circle
  • Area of Circles
  • Conversion of Units
  • Applications in real life.

Apart from practising solutions for chapter 11, Perimeter and Area, students are advised to solve sample papers and previous year question papers, perimeter and area class 7 worksheets to get an idea of the types of questions and also marks contained by this chapter in the final exam. It will also help students to score good marks in the final exam of class 7.

NCERT Solutions For Class 7 Maths Chapter 11 Exercises

Exercise 11.1


Q1. Find

                (a) the area of the land

                (b) the price of the land,

                If the land is of rectangular shape measuring 300 m in length and 400 m in breadth.  The cost of 1m2 land is Rs. 10000.

Ans:       (a) Area = Length (l) x Breadth (b)

= 300 x 400

= 120000 m2

(b) Cost of 1 m2 land = Rs 10000

Cost of 120000 m2 land = 10000 x 120000 = Rs 1200000000.

 

Q2. The perimeter of a park is 240m. Determine the area.

Ans:       Perimeter = 240 m

4 x Length of a side of the park = 240

Length of a side of park = 240/4 = 60 m

Area = (Length of a side of park) 2 = (60)2 = 3600 m2

 

Q3. The area and length of a rectangular plot of land are 1200 m2 and 40 m respectively. Determine its perimeter.  

Ans:       Area = Length (l) x Breadth (b) = 1200 m2

= 40 x Breadth = 1200

= Breadth = 1200/40 = 30

Perimeter = 2 (Length (l) + Breadth (b) )

= 2 (30 + 40) = 2 (70) = 140 m

 

Q4. Find the breadth and area if the perimeter and length of the rectangular sheet is 120 cm and 40 cm respectively.

Ans:       Perimeter = 2 (Length + Breadth) = 120 cm

2 (40 + Breadth) = 120

40 + b = 60

b = 6O – 40 =20 cm

Area = Length (l) x Breadth (b) = 40 x 20 = 800 cm2

 

Q5. The area of the rectangular park is equal to a square park. If the length of rectangular park measures 80m and the side of the square park is 100 m .Determine the breadth of the rectangular park.

Ans:       Area of the square park = (side of the park)2 = (100)2 = 10000 m2

Area of the rectangular park = Length (l) x Breadth (b) = 10000

80 x Breadth = 10000

Breadth = 125 m

 

Q6. The length and breadth of a wire whose shape is a rectangle are 22 and 40 respectively. The wire is then reshaped into a square. Determine the side of the square and find out which shape encloses more area.

Ans:       Perimeter of the rectangle = Perimeter of the square

2 (Length (l) + Breadth (b)) = 4 x Side

2 (22 + 40) = 4 x Side

2 x 62 = 4 x Side

Side =124/4 = 31 cm

Area of the rectangle = 22 x 40 = 880 cm2

Area of the square = (Side)2 = 31 x 31 = 961 cm2

Therefore, the square shaped wire encloses more area.

 

Q7. If the breadth and perimeter of a rectangle are 40 cm and 120 cm respectively, determine the length and area of the rectangle.

Ans:       Perimeter = 2 (Length (l) + Breadth (b)) = 120

2 (Length + 40) = 120

Length + 40 = 60

Length = 60 – 40 = 20 cm

Area = Length (l) x Breadth (b) = 20 x 40 = 800 cm2

 

Q8. A door is fitted in a wall whose length measures 1 m and breadth measures 2 m. The length and breadth of the wall are 5 m and 3.5 m (in the given figure). The price of white washing the wall is Rs 10 per m2. Find the cost of whitewashing the wall without the door.

8

Ans:       Area of the wall = 5 x 3.5 = 17.5 m2

Area of the door = 2x 1 = 2 m2

Area to be white-washed = 17.5 – 2 = 15.5 m2

Cost of white-washing 1 m2 area = Rs 10

Cost of white-washing 15.5 m2 area = 15.5 x 10 = Rs 155

 

Exercise 11.2


Q1. Determine the area of the parallelograms in the diagram given below

Ans:

9

(a) Area of a parallelogram = Height x Base

Base = 8 cm

Height = 5 cm

Area = 8 x 5 = 40 cm2

(b) Area of a parallelogram = Height x Base

Base = 6 cm

Height = 4 cm

Area = 6 x 4 = 24 cm2

(c) Area of a parallelogram = Height x Base

Base = 4.5 cm

Height = 7 cm

Area = 4.5 x 7 = 31.5 cm2

(d) Area of a parallelogram = Height x Base

Base = 9.6 cm

Height = 10 cm

Area = 9.6 x 10 = 96 cm2

(e) Area of a parallelogram = Height x Base

Base = 8.8 cm

Height = 4 cm

Area = 8.8 x 4 = 35.2 cm2

 

Q2. Determine the area of the triangles .

Ans:

10

(a) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 6 cm

Base = 8 cm

Area \(= \frac{1}{2}\times 8 \times 6\) = 48 cm2

(b) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 6.4 cm

Base = 10 cm

Area \(= \frac{1}{2}\times 10 \times 6.4\) = 64 cm2

(c) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 6 cm

Base = 8 cm

Area \(= \frac{1}{2}\times 8 \times 6\) = 48 cm2

(d) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 4 cm

Base = 6 cm

Area \(= \frac{1}{2}\times 4 \times 6\) = 24 cm2

 

Q3. Fill the empty cells:

S. No Base Height Area of parallelogram
A 10 cm 124 cm2
B 12 cm 96 cm2
C 7 cm 91 cm2
D 11 cm 121 cm2

Ans:

(a) Area of a parallelogram = Height x Base

B = 10 cm

H =?

Area = 124 cm2

10 x h = 124

\(h = \frac{124}{10}=12.4\)

So, the height of the parallelogram is 12.4 cm

(b) Area of a parallelogram = Height x Base

B =?

H = 12 cm

Area = 96 cm2

b x 12 = 96

\(b = \frac{96}{12}=8\)

So, the base of the parallelogram is 8 cm

(c) Area of a parallelogram = Height x Base

B = 7 cm

H =?

Area = 91 cm2

7 x h = 91

\(h = \frac{91}{7}=13\)

So, the height of the parallelogram is 13 cm

(d) Area of a parallelogram = Height x Base

B = 11 cm

H =?

Area = 121 cm2

11 x h = 121

\(h = \frac{121}{11}=11\)

So, the height of the parallelogram is 11 cm

 

Q4. Fill in the blanks:    

Base Height Area of triangle
12 24 cm2
30 60 cm2
22 66 cm2

Ans:      

(a) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = ?

Base = 12 cm

Area \(= \frac{1}{2}\times Base \times height\) = 24 cm2

\(\frac{1}{2}\times 12 \times h = 24\) \(h = \frac{24\times 2}{12} = 4 cm\)

Therefore, the height of the triangle is 4 cm

(b) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 30 cm

Base = ?

Area \(= \frac{1}{2}\times Base \times height\) = 60 cm2

\(\frac{1}{2}\times b \times 30 = 60\) \(h = \frac{60\times 2}{30} = 4 cm\)

Therefore, the base of the triangle is 4 cm

(c) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height =?

Base = 22 cm

Area \(= \frac{1}{2}\times Base \times height\) = 66 cm2

\(\frac{1}{2}\times 22 \times h = 24\) \(h = \frac{66\times 2}{22} = 6 cm\)

Therefore, the height of the triangle is 6 cm.

 

Q5. ABCD is a parallelogram (in the given figure). BX is the height from B to CD and BY is the height from B to AD. If CD = 12 cm and BX = 7.6 cm.

Find:      (a) the area of the parallelogram ABCD

                (b) BY, if AD = 8 cm.

Ans:

11

(a) Area of parallelogram = Base x Height = CD x BX

= 7.6 x 12 = 91.2 cm2

(b) Area of parallelogram = Base x Height = AD x BY = 91.2 cm2

BY x 8 = 91.2

BY = 91.2/8 =11.4 cm

 

Q6. SL and QM are the heights on sides PQ and PS respectively of parallelogram PQRS (in the given figure). If area of parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

`Ans:

12

Area of a parallelogram = Height x Base = PQ x SL

1470 = 35 x SL

DL = \(\frac{1470}{35}\) = 42 cm

Also, PS x QM = 1470

1470 = 49 x QM

\(QM = \frac{1470}{35} = 30\ cm\)

 

 Q7. ABC is right angled at A (in the given figure). AD Is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, determine the area of triangle ABC. Also find the length of AD.

Ans:

13

Area of a triangle \(= \frac{1}{2}\times Base \times height\)  \(= \frac{1}{2}\times 5 \times 12\) = 30 cm2

Also, area of triangle \(= \frac{1}{2}\times AD \times BC\) \( 30 = \frac{1}{2}\times AD \times 13\) \(\frac{30\times 2}{13}= AD\)

AD = 4.6 cm

 

Q8. Triangle ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (in the given figure). The height AD from A to BC is 6 cm. Find the area of triangle ABC. What will be the height from C to AB i.e., CE?

Ans:

14

Area of triangle ABC \(= \frac{1}{2}\times Base \times height\) \(= \frac{1}{2}\times BC \times AD\) \(= \frac{1}{2}\times 9 \times 6 = 27 cm2 \)

Area of triangle ABC \(= \frac{1}{2}\times Base \times height\) \(= \frac{1}{2}\times AB \times CE\) \(27= \frac{1}{2}\times 7.5 \times CE\)

CE = 7.2

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