CBSE Class 7 Maths Chapter - 11 Important Questions - Free PDF Download
Very Short Answer Questions (1 – 5) 1 Marks
1. Write the formula to find the area of the triangle.
Ans: The formula to find the area of triangle is
A=$\frac{1}{2}bh$
where, b= Base of triangle
h= Height of triangle
2. Write the formula to find the circumference of a circle.
Ans: The formula to find the circumference of a circle is
C=$2\pi r$
Where, C= Circumference of circle
r= Radius of circle
3. Write the formula to find the area of the circle.
Ans: The formula to find the area of circle is
A=$\pi {{r}^{2}}$
where, A= Area of circle
r= Radius of circle
4. Write the formula to find the area and perimeter of a rectangle.
Ans: The formula to find the area and perimeter of a rectangle are
Area of rectangle=$l\times b$
Perimeter of rectangle=$2(l+b)$
where, l= Length of rectangle
b= Breadth of rectangle
5. The distance around a circular region is known as its_______.
Ans: Circumference
Short Answer Questions (6 – 15) 2 Marks
6. Find the area of a square park whose perimeter is $480$m.
Ans: Given, Perimeter(P)= $480$m
s=Side of square
P= $4\times s$
$480$= $4\times s$
s=$\frac{480}{4}$
s=$120$m
Area of Square=s2
=${{(120)}^{2}}$
=$14,400$sq.m
7. If the perimeter of a rectangle is $390$cm and the length is $30$cm. Find its breadth in the area.
Ans: Given, Perimeter= $390$cm , length=$30$cm
B= Breadth
A= Area
P=$2(l+b)$
$390=2(30+b)$
$\frac{390}{2}=(30+b)$
$195=(30+b)$
$b=195-30$
$b=165$cm
A=$l\times b$
A=$30\times 165$
A=$4,950$sq.cm
8. A wire bent in the shape of a rectangle. Its length is $30$cm and breadth is $15$cm and if the same wire is rebent in the shape of a square. What will be the measure of slides and which encloses more area.
Ans: We know, perimeter of rectangle
P=$2(l+b)$
Where, l=length and b = breadth
=$2(30+15)$
=$2(45)$
=$90$cm
Perimeter of Square= Perimeter of rectangle
$90=4s$ [s=side]
$s=\frac{90}{4}$
$s=22.5$cm
Area of rectangle = $l\times b$
= $30\times 15$
=$450$sq. cm
Area of Square=s2
=${{(22.5)}^{2}}$
=$506.25$sq.cm
Hence, Square encloses a large area.
9. Find the area of
Ans: Given
height=$3$cm
breadth=$6$cm
Area of parallelogram = $b\times h$
= $6\times 3$
= $18$cm2
10. Find the area of
Ans: Given base=$5.5$cm , height=$2.2$cm
Area of triangle = $\frac{1}{2}\times b\times h$
= $\frac{1}{2}\times 5.5\times 2.2$
= $6.05$sq. cm
11. $\Delta ABC$ is isosceles with AB = AC = $5.5$cm and BC = $8$cm. What will be the height from C to AB i.e., CE? If the height AD from A to BC is $4.5$cm. Find the area of $\Delta ABC$?
Ans: Given BC=base=$8$cm , AD=heigth=$4.5$cm
Area of $\Delta ABC$ = $\frac{1}{2}\times b\times h$
= $\frac{1}{2}\times 8\times 4.5$
= $18$sq.cm
Area of $\Delta ABC$ = $\frac{1}{2}\times b\times h$
= $\frac{1}{2}\times AB\times CE$
$18$=$\frac{1}{2}\times 5.5\times CE$
CE=$\frac{36}{5.5}$
CE=$6.55$cm
12. If the circumference of a circular sheet is $132$cm. Find its radius and area.
Ans: Given, Circumference=$132$cm
$C=2\pi r$ [r=radius]
$132=2\times \frac{22}{7}\times r$
$r=\frac{132\times 7}{2\times 22}$
$r=21$cm
Area=$\pi {{r}^{2}}$
Area=$\frac{22}{7}\times {{(21)}^{2}}$
Area=$66\times 21$
Area=$1386$sq. cm
13. Find the circumference and area of the circle if radius is $7$cm.
Ans: Given, radius=$7$cm
$C=2\pi r$ [C=Circumference]
$C=2\times \frac{22}{7}\times 7$
$C=44$cm
Area=$\pi {{r}^{2}}$
Area=\[\frac{22}{7}\times {{(7)}^{2}}\]
Area=$22\times 7$
Area=$154$sq. cm
14. A garden is $50$m long and $42$m broad. A path of $2$m wide is built outside and around it. Find the area of the path in hectares?
Ans: Given,
Length of garden=$50$m
Breadth of garden =$42$m
Area of garden = $l\times b$
= $50\times 42$
=$2100$sq. m
Area of garden where path is included
Area of garden = $l\times b$
= $54\times 46$
=$2484$sq. m
Area of path = Area of garden including path – Area of garden
=$2484-2100$
= $384$sq.m
$1$hectare = $10000$m2
Area of garden= $\frac{384}{10000}$
= $0.0384$hectare
15. Find the area and perimeter of the square whose side is $4$cm.
Ans: We know,
Area of square= ${{s}^{2}}$ [s=side]
=${{4}^{2}}$
=$16$sq. cm
Perimeter = $4s$
=$4\times 4$
=$16$cm
Long Answer Questions (16 – 20) 3 Marks
16. The length and breadth of a rectangular piece of land are $350$m and $150$m respectively.
Find
a. The area
Ans: We know,
Area of garden = $l\times b$
l=length
b= Breadth
= $350\times 150$
=$52,500$sq. m
b. The cost of the land, if $1{{m}^{2}}$ of the land costs Rs. $10,000$
Ans: Given, Cost of $1{{m}^{2}}$of land = Rs. $10,000$
Cost of $52,500{{m}^{2}}$of land = Rs. $52,500\times 10,000$
=Rs. $52,50,00,000$
17. A rectangular field of length $50$m and breadth $45$m need to be fenced. Find the cost of fencing if the changes are Rs. $4$ per meter.
Ans: We know, perimeter of rectangle
P=$2(l+b)$
l=length
b= Breadth
=$2(50+45)$
=$2(95)$
=$190$m
Cost of fencing= $190\times 4$
= Rs. $760$
18. $\Delta ABC$ is right angles at A, AD BC ⊥ .If AB = $8$cm, BC = $17$cm and AC = $15$cm. Find the area of ABC and also length of AD.
Ans: Here,
Area of $\Delta ABC$ = $\frac{1}{2}\times b\times h$
b= base
h=height
=$\frac{1}{2}\times AC\times AB$
= $\frac{1}{2}\times 8\times 15$
= $60$sq.cm
Area of $\Delta ABC$ = $\frac{1}{2}\times b\times h$
= $\frac{1}{2}\times BC\times AD$
$60$=$\frac{1}{2}\times 17\times AD$
AD=$7.03$cm
19. Find the area and circumference of the circle whose radius is
a. $2$cm
Ans: Given, radius=$2$cm
$C=2\pi r$ [C=Circumference]
$C=2\times \frac{22}{7}\times 2$
$C=\frac{88}{7}$cm
Area=$\pi {{r}^{2}}$
Area=\[\frac{22}{7}\times {{(2)}^{2}}\]
Area=$12\frac{4}{7}$sq.cm
b. $21$cm
Ans: Given, radius=$21$cm
$C=2\pi r$
$C=2\times \frac{22}{7}\times 21$
$C=132$cm
Area=$\pi {{r}^{2}}$
Area=\[\frac{22}{7}\times {{(21)}^{2}}\]
Area=$66\times 21$
Area=$1386$sq. cm
20. Find the area of quadrilateral ABCD here AC = $20$cm, BM = $4$cm, DN = $4$cm and BM $\bot $AC and DN $\bot $AC .
Ans: Using formula
Area of $\Delta ABC$ = $\frac{1}{2}\times b\times h$
B= Base
H=Height
= $\frac{1}{2}\times 4\times 20$
= $40$sq.cm
Area of $\Delta ADC$ = $\frac{1}{2}\times b\times h$
=$\frac{1}{2}\times 20\times 4$
= $40$sq.cm
Area of quadrilateral ABCD = Area of $\Delta ABC$+ Area of $\Delta ADC$
=$40+40$
=$80$sq. cm
Area and Perimeter Questions for Class 7 - Free PDF Download
Students looking for chapter 11 class 7 maths extra questions for practice can download the free PDF and start their exam preparations. The PDF also comprises a list of Perimeter and Area Class 7 important questions that are commonly asked in examinations. It also includes solutions to these important questions to help you clear your doubts. You may expect questions from this chapter in your examinations. The formulas for finding the area and perimeter of different geometric figures should be on your fingertips as without them you will not be able to solve the mathematical problem.
The PDF is designed with the objective to impart concept-based knowledge to everyone, free of cost. Mathematics is a subject that requires regular practice. Go through perimeter and area class 7 questions PDF that includes word problems from the chapter. Word problems are a great way to practice these concepts because you have to understand the question and phrase the mathematical equations to solve them.
It is very essential to know your syllabus well and the weightage that each unit in your syllabus carries. It will help you to practice the chapters with the most weightage to improve your score and also help you in bolstering your weak points. To prepare well you must practice as many questions as possible from your textbook to get you familiar with the type of questions that can be asked in the examinations. Practicing your mathematical problems daily will increase your dexterity and problem-solving skills which will also help you in your other subjects.
Class 7 syllabus is divided into six units. The weightage each unit carries is given below. Students can use this information and plan their studies to improve their overall scores.
Class 7 Chapter 11 Weightage Marks
Benefits of Important Questions of Perimeter and Area Class 7
Downloading perimeter and area class 7 important questions will benefit the students immensely. It will prepare them for their examinations and students will get more comfortable while solving problems related to area and perimeter. Students should also practice word problems as it makes their brains work extra in formulating equations rather than just calculating values.
Students can download perimeter and area questions for class 7 here. The benefits of practicing these questions are:
Practicing different types of questions will make students more familiar with the concepts and they will be able to solve questions correctly in exams.
The list of important questions is prepared by the subject experts which have a high probability of coming to your upcoming exams.
These are perfect tools for testing your knowledge of the concepts of the topic.
With enough practice, you can understand the pattern of questions and how to solve them.
Word problems can be tricky to solve so it is important that you are well versed in how to solve them.
Conclusion
Perimeter and Area is an integral part of Class 7 Maths and plays a crucial role from an examination perspective. The important questions for Class 7 Maths, as discussed by NCERT, cover a wide range of topics within the subject. They also provide a concise guide to critical points and details related to the topic.
A solid understanding of each section of Class 7 Maths is fundamental as it forms the basis for higher-level studies. However, this section primarily focuses on important questions within the context of Class 7 Maths.
FAQs on Important Questions for CBSE Class 7 Maths Chapter 11 - Perimeter and Area
1. List Down the Important Formulas that are Required to Solve Mathematical Problems from the Chapter - Area and Perimeter.
The formulas of Area and Perimeter of some standard polygons are given below. Students should make a habit of practicing questions that will automatically allow them to retain the formulas in their examinations.
Given below are the important formulas for some standard polygons.
For a rectangle of length “l” and breadth “b”.
Area of rectangle= l X b
Perimeter of rectangle = 2(l+b)
For a square of the side “a”,
Area of Square=a2
Perimeter of square=4a
For a circle with radius “r”
Area of Circle = πr2
Circumference of Circle=2πr
For a triangle of base “b” and height “h”
Area of triangle=1/2 X b X h
Perimeter of triangle= Sum of all sides
For an equilateral triangle of the side “a”
Area of an equilateral triangle = √3a2/4
Perimeter of an equilateral triangle = 3a
2. How can I Benefit from Practicing Maths Important Questions?
The examination pattern is divided into four sections- Section A, B, C, and D. To solve questions from each section you need to practice a lot of questions regularly. Practicing important questions will make you familiar with the type of questions. If you are well versed with the concepts you will be easily able to solve direct questions.
To solve word problems you will need to convert them into solvable equations first before you attempt them. Practicing important questions will give you the confidence to solve any difficult question that comes in the examination.
One great way to practice important questions from each chapter is to solve sample papers. If you solve sample papers under the stipulated time it will benefit you because you will be familiar with working under time pressure. Focus on understanding the concept of the important questions and the pattern they follow.
3. What is the difference between perimeter and area?
The perimeter of a closed figure is the distance around it, whereas its area is the fraction of a plane or region it encloses. A regular polygon's perimeter is equal to the number of sides multiplied by the length of one side. Similarly, the perimeter and area of various forms vary and are calculated using different formulae based on this concept. Perimeter and area are important concepts of maths and have significant weightage in the exam. You can prepare them well with Vedantu.
4. How can I download the PDF file for the Solutions of NCERT Class 7 Maths Chapter 11?
The solutions are easily available on the Vedantu site.
Click on this link
The webpage with Vedantu’s solutions for Class 7 Maths Chapter 11 will open.
To download this, click on the Download PDF button and you can view the solutions offline.
The Vedantu website has a variety of instructors that can aid you with this or any other chapter.
5. What is an irregular shape Class 7?
Any size and length is a possibility for an irregular form. A kite, a diamond shape, a leaf, and other irregular forms may be seen all around us. An irregular form is defined as one whose sides and angles are not of equal length. Rocks, prisms, beans, and screws are examples of irregular solids. Each of these things has a unique irregular form that isn't made up of regular shapes. The difficulty to estimate an irregular solid area, volume, or mass by measuring its length, breadth, and height distinguishes it.
6. What is the difference between irregular and regular shapes?
An irregular form can be any size and length, whereas a regular shape has equal sides and angles. Thus, the area of a regular shape may be calculated directly using appropriate formulae, but the area of an irregular shape can be calculated by decomposing an irregular shape into numerous regular ones. Irregular shapes’ areas are generally harder to calculate due to their unpredictability. Hence, an approach different from the norm, while calculating the area of regular shapes, has to be taken for irregular shapes.
7. How do you find areas of irregular shapes?
To calculate the area of an irregular form, first split it into regular shapes that you can identify, such as triangles, rectangles, circles, squares, and so on. Then, to create an area of irregular forms, determine the area of each individual shape and put them together. Due to their unpredictability, the areas of irregular forms are typically more difficult to compute. For more problems regarding this, you can visit the Vedantu site.