# Important Questions Class 7 Mathematics Chapter 11

## Important Questions Class 7 Mathematics Chapter 11 – Perimeter and Area

Mathematics is a vital subject students study in school. Mathematics is needed in every aspect of our lives to solve real-life problems. Students learned what an area is in their earlier classes. In this chapter, they will learn about perimeter too.

Class 7 Mathematics Chapter 11, discusses the area and perimeter of geometric shapes. Area means the space occupied by a geometric shape, while the perimetre is the length of all sides of any geometric shape. In this chapter, students will study how to calculate a rectangle’s and square’s area and perimeter. Students must practice questions to build their concepts and improve their preparation for the exams.

Extramarks is a reputed educational company in India. Our experts have made the Important Questions Class 7 Mathematics Chapter 11 to help students in practice. They have collected the questions from different sources, such as the textbook exercises, CBSE sample papers, CBSE past years’ question papers, NCERT exemplar and important reference books. They have also solved the questions, and experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Important Questions Class 7 Mathematics Chapter 11 will help students  score better in exams.

Extramarks provides all the important study materials related to CBSE and NCERT, and you can download the study materials after registering on our official website. We provide CBSE syllabus, CBSE sample papers, CBSE past years’ question papers, CBSE revision notes, CBSE extra questions, NCERT books, NCERT solutions, NCERT important questions, NCERT exemplar, vital formulas, and many more.

Important Questions Class 7 Mathematics Chapter 11 – With Solutions

The experts at Extramarks have made the question papers by collecting questions from different sources. They have taken help from the textbook exercises, CBSE sample papers, CBSE past years’ question papers, and important reference books. They have answered the questions, and experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Important Questions Class 7 Mathematics Chapter 11 will help students  score better in exams. The important questions are-

Question 1. A rectangular piece of land has a length of 500 m and a breadth of 300 m respectively. Find the area and the cost of the land. The condition given is 1 m2 of the land cost Rs 10,000.

According to the given question,

The length of the triangle = 500 m

The breadth of the triangle = 300 m

Since the formula for area = length × breadth = 500 × 300 m2 = 150000 m2

If Rs 10,000 is the cost of land of 1m2

Then the cost of 150000 m2 land = 10000 × 150000 = Rs 1500000000

Question 2. The perimetre is given at 320 m. Calculate the area of the square park.

Answer 2: According to the given question,

the Perimeter of the square = 320 m

Length of the side of the park = 320/4 = 80 m

Since 4 × length of the side of the park = 320 m

Area of the square park = (length of the park side)2 = (80)2 = 6400 m2.

Question 3. Calculate the breadth and perimetre of the rectangular plot when the area given is 400 m2 and the length given is 22 m.

According to the given question,

The Area of the given rectangular plot = 440 m2

Length of the plot = 22 m

Applying the formula, area = length × breadth

Breadth = 440/22 = 20 m

Perimeter of the rectangle will be calculated as = 2(length + Breadth) = 2(22 + 20) = 2 × 42 = 84 m.

So the rectangular plot has a perimetre of 84 m

Question 4. Calculate the breadth when the perimetre and length of the rectangular sheet given are 100 cm and 35 cm respectively. Also, calculate the area.

According to the given question,

Perimetre given is 100 m

Length of the rectangular sheet given is 35 cm

Since perimetre = 2 ( length + breadth )

On applying the formula

100 = 2 (35 + breadth)

Therefore, breath is calculated as 15 cm.

Since we know that the

Area = length × breadth = 35 × 15 = 525 cm2

So, the area of the rectangular sheet is 525 cm2

Question 5. The square park and the rectangular park have the same area. Calculate the breadth of the rectangular park when the side of the rectangular park is 60 m and the length of the rectangular park is 90 m.

According to the given question,

the Area of the square park = area of the rectangular park

Side of the square park is 60 m

Length of the rectangular park is 90 m

We know the formula that,

Area of the square park = (one of the sides of the square )2 = (60)2 = 3600 m2

Area of the rectangular park = 3600 m

On applying the formula

Question 6. Suppose the shape of the wire is a rectangle. The length and breadth of the rectangle are 40 cm and 22 cm respectively. What is going to be the measurements of each side if the wire is bent in the shape of a square? Find out about the shape which encloses more area.

The perimetre of the square = perimetre of a rectangle, as given in the question

According to the given question,

the Length of the rectangle = 40 cm

Breadth of the rectangle = 22 cm

Since the perimetre of square and rectangle are equal

2 (length + breadth) = 4 × side

2(40 + 22) = 4 × side

2 × 62 = 4 × side

124 = 4 × side

124/4 = side = 31 cm

Area of the rectangle = length × breadth = 40 × 22 = 880 cm2

Area of the square = side2 = 31 × 31 = 961 cm2

Based on the above values it could be said that the square shape enclosed more area.

Question 7. The length of the door is 2 m and the breadth is 1 m and it is fitted in a wall. The wall has a length of 4.5 m and a breadth of 3.6 m, what is the cost required for whitewashing the wall when the rate is Rs 20 per m2 for whitewashing.

According to the given question,

The Door has length = 2 m

Door has breadth = 1 m

Wall has length = 4.5 m

Wall has breadth = 3.6 m

Area of the door = length × breadth = 2 × 1 = 2 m2

Area of the wall = length × breadth = 4.5 × 3.6 = 16.2 m2

The area for whitewashing = 16.2 – 2 = 14.2 m2

The cost of the whitewashing 1 m2 area = 20 Rs

Cost of whitewashing the area of 14.2 m2 = 14.2 × 20 = 284 Rs

Question 8. The parallelograms are given below to calculate the area.

1. As given in the figure,

Height = 4 cm, base = 7 cm

Area of the parallelogram = base × height = 7 × 4 = 28 cm2

1. Height = 3 cm, base = 5 cm

Area of the parallelogram = base × height = 5 × 3 = 15 cm2

1. Height = 3.5 cm, base = 2.5 cm

Area = 2.5 × 3.5 = 8.75 cm2

1. Height = 4.8 cm, base = 5 cm

Area = 5 × 4.8 = 24 cm2

1. Height = 4.4 cm, base =2 cm

Area of the parallelogram = 2 × 4.4 = 8.8 cm2

Question 9. Find the area of the triangle given below.

3.

4.

1. As per the given figure,

The Base of the triangle = 4 cm

Height of the triangle = 3 cm

Area of the triangle = ½ × base × height = ½ × 4 × 3 = 6 cm2

1. According to the given figure,

The Base of the triangle = 3.2 cm

Height of the triangle = 5 cm

Area of the triangle = ½ × base × height = ½ × 3.2 × 5 = 8 cm2

1. According to the given figure,

The Base of the triangle = 3 cm

Height of the triangle = 4 cm

Area of the triangle = ½ × base × height = ½ × 3 × 4 = 6 cm2

1. According to the given figure,

The Base of the triangle = 3 cm

Height of the triangle = 2 cm

Area of the triangle = ½ × base × height = ½ × 3 × 2 = 3 cm2

Question 10. PQRS is a parallelogram, 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 then calculate the area of the parallelogram PQRS. Calculate QN when PS gave as 8 cm.

SR = 12 cm, QM = 7.6 cm

Area of the parallelogram = length × breadth = SR × QM = 12 × 7.6 = 91.2 cm2

For the second part of the question, calculating QN includes:

Area of the parallelogram = length × breadth

91.2 = PS × QN

91.2 = 8 × QN

QN = 91.2/8 = 11.4 cm

Question 11. DL and BM are the heights on sides AB and AD respectively of ABCE parallelogram. Let us suppose the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm. Calculate the length of the BM and the DL.

According to the question given above,

The Area of the parallelogram = 1470 cm2

AB = 35 cm

Then

It is clear that

The Area of the parallelogram = base × height

1470 = AB × BM

1470 = 35 × DL

1470/35 = DL

DL = 42 cm

Area of the parallelogram = base × height

1470 = 49 × BM

BM = 1470/49

BM = 30 cm

Question 12. The right-angled triangle is ABC given. This is right-angled at A. AD is perpendicular to BC. If AB = 5 cm and BC = 13 cm and AC = 12 cm. Calculate the area of the ABC. Also, calculate the length of the AD.

According to the given question,

AB = 5 cm

BC = 13 cm

AC = 12 cm

We know that

Area of the triangle ABC = ½ × Base × height

= ½ × AB × AC

=½ × 5 × 12 = 1 × 5 × 6 = 30 cm2

Now,

Area of the triangle ABC = ½ × Base × height

30 = ½ × AD × BC

30 = ½ × AD × 13

AD = 60/13 = 4.6 cm

Question 13. Find the circumference of the circle with a radius of 14 cm. take the value of =22/7

Solution 13.

As per the question given, the radius of the circle = 14 cm

Circumference of the circle will be calculated by the formula = 2

= 2 × 22/7 ×14 = 2 × 22 × 2 = 88 cm

Question 14. The circumference of a circular sheet is 154 m. calculate the radius. Calculate the area of the sheet as well.

As per the question, it is given that

The Circumference of the circle = 154 m

We know the formula

Circumference of the circle will be calculated by the formula = 2

154 = 2 × 22/7 × r

154 = 44/7 × r

r = 154 × 7/ 44

r = 14 × 7/ 4

r = 7 × 7/ 2

r = 49/2 = 24.5 m

area of the circle = r2 = 22/7 × (24.5)= 22/7 × 600.25

= 22 × 85.75 = 1886.5 m2

Question 15. Calculate the perimetre of the figure. The figure is a semicircle including its diameter.

As per the given question,

The Diameter of the semi-circle = 10 cm

Radius = r = d/2 = 10/2 = 5 cm

Circumference of the semi-circle = 22/7 × 5 = 110/7 = 15.71 cm

To calculate the perimeter of the above figure,

Perimeter of the semi-circle = semi-circle circumference + semicircle diameter

= 15.71 + 10 = 25.71 cm

Question 16. The rate of polishing is Rs 15/m2. Find the cost of polishing a circular table – top which has a diameter of 1.6 m.

Answer 16: As per the given question,

The diameter of the circular table – top = 1.6 m

As we know the

Radius = r = d/2 = 1.6 / 2 = 0.8 m

Area of the circular table top will be 3.14 × 0.8 × 0.8 = 2.0096 m2

The cost of polishing 1 m2 area = Rs 15

So for calculating the 2.0096 m2 area = Rs 15 × 2.0096 = 30.144

So in order to polish the area of 2.0096 m2, the cost incurred is Rs 30.144

Question 17. Fill in the blanks:

• ________ is the distance around a closed figure.
• _________ is the part of the plane which is occupied by the closed figure.
• The formula to calculate the perimetre of the square will be presented as  _______
• The formula to calculate the perimetre of the rectangle will be presented as _______
• The area of the square is calculated as ________
• The area of a rectangle is calculated as ________
• The area of the parallelogram is calculated as _______
• ___________ is the distance around a circular region.

1. Perimeter is the distance around a closed figure.
2. Area  is the part of the plane which is occupied by the closed figure.
3. The formula to calculate the perimetre of the square will be presented as  4 × side
4. The formula to calculate the perimetre of the rectangle will be presented as  2 × (length + breadth)
5. The area of the square is calculated as side × side
6. The area of a rectangle is calculated as length × breadth
7. The area of the parallelogram is calculated as base × height
8. Circumference is the distance around a circular region.

Question 18. Find the area of the quadrilateral ABCD. See the figure given below.

Answer 18. As shown in the figure,

AC = 22 cm, BM = 3 cm

DN = 3 cm

BM is perpendicular to AC

DN is perpendicular to AC.

It is clear from the figure that in order to calculate the area of the quadrilateral ABCD, the first step is the calculation of the area of the triangle ABC and ADC.

To calculate the area of the triangle ABC,

= ½ × base × height

= 1 × 11 × 3 = 33 cm2

To calculate the area of the triangle ADC,

= ½ × base × height

= ½ × 22 × 3 = 33 cm2

Area of the quadrilateral = area of ABC + area of ADC

= 33 + 33 = 66 cm2

Question 19. Calculate the area of the shaded portion.

To find the area of EFDC, the area of triangle AEF, EBC and Rectangle ABCD has to be calculated.

To calculate the area we will apply the formula, ½ × base × height

Area of AEF = ½ × 6 × 10  = 30 cm2

Area of EBC = ½ × 8 × 10  = 40 cm2

Area of rectangle ABCD = length × breadth = 18 × 10 = 180 cm2

Area of EFDC will be calculated as,

ABCD area – (AEF – EBC)

= 180 – (30 + 40)

= 180 – 70

= 110 cm2

Question 20. Look at the figure given below, the circular card sheet has a radius of 14 cm. Two circles are removed of radius 3.5 cm and a rectangle is also removed, the length and breadth of the rectangle are 3 cm and 1cm respectively. Given the value = 22/7, calculate the area of the remaining sheet by applying the suitable formulas.

From the question, it is clear that

The Radius of the circular sheet is 14 cm

The Radius of the two small circles is 3.5 cm

Length of the rectangle is 3 cm

Breadth of the rectangle is 1 cm

To calculate the remaining area,

The Area of the circular card sheet = 22/4 × 14 × 14 = 22 × 2 × 14 = 616 cm2

To calculate the area of the two small circles,

= 2 × (22/7 × 3.5 × 3.5)

=2 × ((22/7) 12.25)

= 2 × 38.5

= 77 cm2

To calculate the area of the rectangle = length × breadth = 3 × 1 = 3 cm2

To calculate the area of the remaining part,

Area of card sheet – ( area of 2 small circles + rectangle area)

= 616 – (77 + 3)

= 616 – 80

= 536 cm2

### Benefits of Solving Important Questions Class 7 Mathematics Chapter 11

Practice is very important for students because it helps them  boost their confidence and clarify their doubts. Students must practise as much as possible to score better in the exams. The experts at Extramarks have made this question series to help them prepare. There will be multiple benefits to solving the questions. The benefits of solving the Important Questions Class 7 Mathematics Chapter 11 are-

• The experts have collected the questions from different sources, such as the textbook exercises, CBSE sample papers, CBSE past years’ question papers, important reference books and NCERT exemplar. Thus, students don’t have to search for questions in different sources,  they will find them in a single pdf. Thus, the Class 7 Mathematics Chapter 11 Important Questions will help students solve more questions and score better in exams.
• The experts have not only collected the questions but have also solved them.They followed CBSE guidelines while writing the answers. Furthermore, they have provided sufficient diagrams to explain each question. Experienced professionals have checked the solutions to ensure the best quality of the writing. Students can follow the answers if they have any doubts. So, the Mathematics Class 7 Chapter 11 Important Questions will help students  score better in exams.
• The experts have tried to incorporate as many questions as possible. They have ensured that every important concept is included in the question series. Many students tend to fear mathematics because they have problems understanding the question. Practice can help them a lot by boosting their confidence. Thus, the Chapter 11 Class 7 Mathematics Important Questions will help students better understand and build their concepts.

Extramarks is a leading company that provides all the important study materials related to CBSE and NCERT. You can download these study materials if you register on our official website. We provide CBSE syllabus, CBSE sample papers, CBSE past years’ question papers, CBSE revision notes, CBSE extra questions, NCERT books, NCERT solutions, NCERT important questions, NCERT Exemplars, vital formulas, and many more. Like the Important Questions Class 7 Mathematics Chapter 11, you will also find important questions for other chapters. Links to the important questions are-

• NCERT books
• Important questions
• CBSE Revision Notes
• CBSE syllabus
• CBSE sample papers
• CBSE past years’ question papers
• Important formulas
• CBSE extra questions