NCERT Solutions for Class 7 Maths Chapter 9 Perimeter and Area Exercise 9.1 - FREE PDF Download
NCERT solutions for Perimeter and Area Class 7 Exercise 9.1 will serve as your companion as you navigate Exercise 9.1. We will break down the key topics that students will encounter, including calculating circle properties like radius and area, tackling practical applications of circles, and mastering word problems that involve these fascinating shapes. By working through class 7 maths exercise 9.1 solutions, you will transform your theoretical knowledge of circles into a powerful tool for solving everyday problems!
Glance on NCERT Solutions Maths Chapter 9 Exercise 9.1 Class 7 | Vedantu
Chapter 9 of Class 7 Maths in the NCERT curriculum focuses on the concepts of Perimeter and Area.
Exercise 9.1 focuses on calculating the perimeter and area of various geometrical shapes such as rectangles, squares, triangles, and parallelograms.
There are 8 questions in Maths Chapter 9 Class 7th Exercise 9.1 which experts at Vedantu fully solved.
Many questions are accompanied by diagrams to help students visualize the problems better.
Formulas Used in Maths Class 7 Chapter 9 Exercise 9.1
The formulas for calculating the area = πr²
Circumference of circle = 2πr
Exercise 9.1
1. Find the area of each of the following parallelograms:
(a)
Ans: Height of parallelogram is \[4cm\] .
Base of the parallelogram is \[7cm\] .
Therefore,
Area of the parallelogram is \[base \times height\] .
The area of parallelogram is \[7cm \times 4\;cm\].
So, the area of parallelogram is \[28c{m^2}\] .
(b)
Ans: Height of parallelogram is \[3cm\] .
Base of the parallelogram is \[5cm\] .
Therefore,
Area of the parallelogram is\[base \times height\] .
Area of parallelogram is \[5cm \times 3cm\]
Area of the parallelogram is \[15c{m^2}\] .
(c)
Ans: Height of parallelogram is \[3.5cm\] .
Base of parallelogram is \[2.5cm\] .
Therefore,
Area of parallelogram is \[base \times height\]
Area of parallelogram is \[2.5cm \times 3.5cm\]
Area of parallelogram is \[8.75c{m^2}\]
(d)
Ans: Height of parallelogram is \[4.8cm\].
Base of the parallelogram is \[5cm\].
Therefore,
Area of the parallelogram is \[base \times height\] .
Area of the parallelogram is \[4.8cm \times 5cm\] .
Area of the parallelogram is \[24c{m^2}\] .
(e)
Ans: Height of parallelogram is \[4.4cm\] .
Base of the parallelogram is \[2cm\] .
Therefore,
Area of the parallelogram is \[base \times height\] .
Area of the parallelogram is \[4.4cm \times 2cm\] .
Area of the parallelogram is \[8.8c{m^2}\] .
2. Find the area of each of the following triangles:
(a)
Ans: Base of the triangle is \[4cm\] .
Height of height is \[3cm\] .
Therefore,
Area of the triangle is \[\dfrac{1}{2} \times base \times height\]..
Area of the triangle is \[\dfrac{1}{2} \times 4cm \times 3cm\] .
Area of the triangle is \[6c{m^2}\].
(b)
Ans: Base of triangle is \[5cm\] .
Height of height is \[3.2cm\] .
Therefore,
Area of the triangle is \[\dfrac{1}{2} \times base \times height\].
Area of the triangle is \[\dfrac{1}{2} \times 5cm \times 3.2cm\] .
Area of the triangle is \[8c{m^2}\] .
(c)
Ans: Base of triangle is \[4cm\] .
Height of height is \[3cm\] .
Therefore,
Area of the triangle is \[\dfrac{1}{2} \times base \times height\].
Area of the triangle is \[\dfrac{1}{2} \times 4cm \times 3cm\] .
Area of the triangle is \[6c{m^2}\] .
(d)
Ans: Base of triangle is \[3cm\] .
Height of height is \[2cm\] .
Therefore,
Area of the triangle is \[\dfrac{1}{2} \times base \times height\].
Area of the triangle is \[\dfrac{1}{2} \times 3cm \times 2cm\] .
Area of the triangle is \[3c{m^2}\] .
3. Find the missing values:
Ans (a) : Base of parallelogram is \[20cm\] .
Height of the parallelogram is to be known.
Area of the parallelogram is \[246c{m^2}\] .
Therefore,
\[Area = base \times height\]
\[246c{m^2} = 20cm \times height\]
\[\dfrac{{246c{m^2}}}{{20cm}} = height\]
\[12.3cm = height\]
Therefore, the height of the parallelogram is \[12.3cm\] .
Ans (b) : Base of parallelogram is to be known .
Height of the parallelogram is \[15cm\] .
Area of the parallelogram is \[154.5c{m^2}\] .
Therefore,
\[Area = base \times height\]
\[154.5c{m^2} = base \times 15cm\]
\[\dfrac{{154.5c{m^2}}}{{15cm}} = base\]
\[10.3cm = base\]
Therefore, the base of the parallelogram is \[10.3cm\] .
Ans (c) : Base of parallelogram is to be known .
Height of the parallelogram is \[8.4cm\] .
Area of the parallelogram is \[48.72c{m^2}\] .
Therefore,
\[Area = base \times height\]
\[48.4c{m^2} = base \times 8.4cm\]
\[\dfrac{{48.4c{m^2}}}{{8.4cm}} = base\]
\[5.8cm = base\]
Therefore, the base of the parallelogram is \[5.8cm\] .
Ans (d) : Base of parallelogram is \[15.6cm\] .
Height of the parallelogram is to be known.
Area of the parallelogram is \[16.38c{m^2}\] .
Therefore,
\[Area = base \times height\]
\[16.38c{m^2} = 15.6cm \times height\]
\[\dfrac{{16.38c{m^2}}}{{15.6cm}} = height\]
\[1.05cm = height\]
Therefore, the height of the parallelogram is \[1.05cm\] .
So, the final table is
4. Find the missing values:
Ans (a) : Base of triangle is \[15cm\] .
Height of the triangle is to be known.
Area of the triangle is \[87c{m^2}\] .
Therefore,
\[Area = \dfrac{1}{2} \times base \times height\]
\[87c{m^2} = \dfrac{1}{2} \times 15cm \times height\]
\[\dfrac{{87c{m^2} \times 2}}{{15cm}} = height\]
\[11.6cm = height\]
Therefore, the height of the triangle is \[11.6cm\] .
Ans (b) : Base of triangle is to be known.
Height of the triangle is \[31.4mm\].
Area of the triangle is \[1256m{m^2}\] .
Therefore,
\[Area = \dfrac{1}{2} \times base \times height\]
\[1256m{m^2} = \dfrac{1}{2} \times base \times 31.4mm\]
\[\dfrac{{1256m{m^2} \times 2}}{{31.4m}} = base\]
\[80mm = base\]
Therefore, the base of the triangle is \[80mm\] .
Ans (c) : Base of triangle is \[22cm\] .
Height of the triangle is to be known.
Area of the triangle is \[170.5c{m^2}\] .
Therefore,
\[Area = \dfrac{1}{2} \times base \times height\]
\[170.5c{m^2} = \dfrac{1}{2} \times 22cm \times height\]
\[\dfrac{{170.5c{m^2} \times 2}}{{22cm}} = height\]
\[15.5cm = height\]
Therefore, the height of the triangle is \[15.5cm\] .
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.
Fig \[11.3\]
Find:
(a) The area of the parallelogram PQRS
Ans: Base of parallelogram is \[12cm\] .
Height of the parallelogram is \[7.6cm\] .
Therefore,
Area of parallelogram is \[base \times height\]
Area of the parallelogram is \[SR \times QM\] .
The area of parallelogram \[12cm \times 7.6cm\;\]
So, the area of parallelogram is \[91.2c{m^2}\].
(b) QN, if PS = 8 cm
Ans: The area of parallelogram is \[91.2c{m^2}\] .
Base of the parallelogram is \[8cm\] .
Height of the parallelogram is to be known.
Area of the parallelogram is \[base \times height\] .
\[Area = base \times height\]
\[Area = PS \times QN\]
\[91.2c{m^2} = 8cm \times QN\]
\[QN = \dfrac{{91.2c{m^2}}}{{8cm}}\]
\[QN = 11.4cm\].
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 \[1470c{m^2}\], \[AB = 35cm\]cm and \[AD = 49cm\], find the length of BM and DL.
Ans: Case 1 : Finding the value of DL.
Area of parallelogram is ABCD \[1470c{m^2}\] .
Base of parallelogram is ,
\[AB = 35cm\]
Height of parallelogram is,
\[DL = ?\]
\[Area = base \times height\]
\[Area = AB \times DL\]
\[1470c{m^2} = 35cm \times DL\]
\[DL = \dfrac{{1470c{m^2}}}{{35cm}}\]
\[DL = 42cm\]
Therefore, the value of DL is \[42cm\] .
Case 2: Finding the value BM.
Case 1 : Finding the value of DL.
Area of parallelogram is ABCD \[1470c{m^2}\] .
Base of parallelogram is ,
\[AD = 49cm\]
Height of parallelogram is,
\[BM = ?\]
\[Area = base \times height\]
\[Area = AD \times BM\]
\[1470c{m^2} = 49cm \times BM\]
\[BM = \dfrac{{1470c{m^2}}}{{49cm}}\]
\[BM = 30cm\]
Therefore, the value of DL is \[30cm\] .
7. \[\Delta ABC\] is right angled at A (Fig 11.25). AD is perpendicular to BC. If\[AB = 5cm\], \[BC = 13cm\] cm and \[AC = 12cm\], Find the area of \[\Delta ABC\]. Also find the length of AD.
Ans: Finding the area of \[\Delta ABC\] .
Base of \[\Delta ABC\] is \[5cm\] .
Height of \[\Delta ABC\] is \[12cm\] .
Therefore,
\[Area = \dfrac{1}{2} \times base \times height\]
\[Area = \dfrac{1}{2} \times AB \times AC\]
\[Area = \dfrac{1}{2} \times 5cm \times 12cm\]
\[Area = 30c{m^2}\]
Therefore, the area of \[\Delta ABC\] is \[30c{m^2}\] .
Finding the value of AD.
Area of \[\Delta ABC\] is \[30c{m^2}\]
Base of \[\Delta ABC\] is \[13cm\] .
Height of \[\Delta ABC\] is to be known.
Therefore,
\[Area = \dfrac{1}{2} \times base \times height\]
\[Area = \dfrac{1}{2} \times BC \times AD\]
\[30c{m^2} = \dfrac{1}{2} \times 13 \times AD\]
\[\dfrac{{30c{m^2} \times 2}}{{13cm}} = AD\]
\[AD = 4.6cm\]
Therefore, the value of AD is \[4.6cm\] .
8. \[\Delta ABC\] 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 ΔABC. What will be the height from C to AB i.e., CE?
Ans: Finding the area of \[\Delta ABC\] .
Base of \[\Delta ABC\] is \[9cm\] .
Height of \[\Delta ABC\] is \[6cm\] .
Therefore,
\[Area = \dfrac{1}{2} \times base \times height\]
\[Area = \dfrac{1}{2} \times AD \times BC\]
\[Area = \dfrac{1}{2} \times 9cm \times 6cm\]
\[Area = 27c{m^2}\]
Therefore, the area of \[\Delta ABC\] is \[27c{m^2}\] .
Find the value of CE.
Area of \[\Delta ABC\] is \[27c{m^2}\]
Base of \[\Delta ABC\] is \[7.5cm\] .
Height of \[\Delta ABC\] is to be known.
Therefore,
\[Area = \dfrac{1}{2} \times base \times height\]
\[Area = \dfrac{1}{2} \times BA \times CE\]
\[27c{m^2} = \dfrac{1}{2} \times 7.5cm \times CE\]
\[\dfrac{{27c{m^2} \times 2}}{{7.5cm}} = CE\]
\[CE = 7.2cm\]
Therefore, the value of CE is \[7.2cm\].
Conclusion
NCERT Solutions for Class 7 Maths Exercise 9.1 Solutions specifically target strengthening your understanding of perimeters and areas of various shapes, with a focus on rectangles and squares. You will revisit familiar concepts like length, breadth, and area, but with an emphasis on solving real-world problems. Sharpen your skills in calculating these properties using the formulas (perimeter = 2(l + b) for rectangles and perimeter = 4a for squares). The real challenge lies in applying these concepts practically. Imagine determining the length of the fence required for a rectangular garden, or the amount of material needed to cover a square floor.
Class 7 Maths Chapter 9: Exercises Breakdown
CBSE Class 7 Maths Chapter 9 Other Study Materials
Chapter-Specific NCERT Solutions for Class 7 Maths
Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
FAQs on NCERT Solutions for Class 7 Maths Chapter 9 - Perimeter and Area Exercise 9.1
1. What is the main focus of Chapter 11 Exercise 11.2 in Class 7 Maths?
The main focus of Exercise 11.2 is to strengthen students' understanding of calculating the perimeter and area of rectangles and squares. It emphasizes practical applications of these concepts through word problems.
2. What formulas are used in Exercise 11.2 for calculating the perimeter and area of rectangles and squares?
The formulas used are:
The perimeter of a rectangle: Perimeter=2(l+b)Perimeter=2(l+b), where ll is the length and bb is the breadth.
The perimeter of a square: Perimeter=4aPerimeter=4a, where aa is the side length.
Area of a rectangle: Area=l×bArea=l×b.
Area of a square: Area=a2Area=a2.
3. Why is it important to learn about the perimeter and area of shapes?
Learning about the perimeter and area of shapes is important because these concepts are widely used in real life. They help in solving practical problems like determining the amount of material needed for construction, fencing a garden, or tiling a floor.
4. How can I apply the concepts learned in Exercise 11.2 to real-world problems?
You can apply these concepts by using the formulas to solve problems related to real-life situations. For example, you can calculate the length of fencing needed for a garden, the area of a room to be carpeted, or the amount of paint required for a wall.