Class 7 maths chapter 11 exercise 11.2 solutions

NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.2 - Perimeter and Area In this exercise, students will calculate the area of parallelograms. They will solve sums based on the formula of Area of the parallelogram = base × altitude and area of triangle from parallelograms which is Area of the triangle = ½ × b × h. Exercise 11.2 Question 1. Find the area of each of the following parallelograms: Solution (a) Height of parallelogram = 4 cm Base of parallelogram = 7 cm Then, Area of parallelogram = base × height = 7 × 4 = 28 cm 2 (b) Height of parallelogram = 3 cm Base of parallelogram = 5 cm Then, Area of parallelogram = base × height = 5 × 3 = 15 cm 2 (c) Height of parallelogram = 3.5 cm Base of parallelogram = 2.5 cm Then, Area of parallelogram = base × height = 2.5 × 3.5 = 8.75 cm 2 (d) Height of parallelogram = 4.8 cm Base of parallelogram = 5 cm Then, Area of parallelogram = base × height = 5 × 4.8 = 24 cm 2 (e) Height of parallelogram = 4.4 cm Base of parallelogram = 2 cm Then, Area of parallelogram = base × height = 2 × 4.4 = 8.8 cm 2 Question 2. Find the area of each of the following triangles: Solution (a) Base of triangle = 4 cm Height of height = 3 cm Then, area of triangle = ½ × base × height = ½ × 4 × 3 = 6 cm 2 (b) Base of triangle = 5 cm Height of height = 3.2 cm Then, area of triangle = ½ × base × height = ½ × 5 × 3.2 = 8 cm 2 (c) Base of triangle = 3 cm Height of height = 3 cm Then, area of triangle = ½ × base × height = ½ × 3 × 4 = 6 cm 2 (d) Base of ...

The main objective of providing exercise-wise solutions in PDF is to help students with their exam preparation. The questions present in this exercise have been solved by BYJU’S experts in Maths, and these solutions guide students to study effectively. The students can download solutions that are available in PDF format easily. Chapter 11 Percentage Exercise 11.2 are provided here. This exercise contains the step-wise method to the conversion of a ratio into per cent and vice versa. Previous Next Access answers to Maths RD Sharma Solutions for Class 7 Chapter 11 – Percentage Exercise 11.2 Exercise 11.2 Page No: 11.4 1. Express each of the following ratios as per cents: (i) 4: 5 (ii) 1: 5 (iii) 11: 125 Solution: (i) Given 4: 5 4: 5 can be written as (4/5) = (4/5) × 100 = 80% (ii) Given 1: 5 1: 5 can be written as (1/5) = (1/5) × 100 = 20% (iii) Given 11: 125 11: 125 can be written as (11/125) = (11/125) × 100 = (44/5) % 2. Express each of the following percents as ratios in the simplest form: (i) 2.5% (ii) 0.4% (iii) 13 3/4 % Solution: (i) Given 2.5% = (2.5/100) = (25/1000) = (1/40) (ii) Given 0.4% = (0.4/100) = (4/1000) = (1/250) (iii) Given 13 3/4 % 13 3/4 = 13.75 = 13.75/100 = 1375/10000 = 11/80

Free PDF download of NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.2 (EX 11.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 7 Science , Maths solutions and solutions of other subjects. 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 \[8.8c\] . 2. Find the area of each of the following triangles: (a) S.No. Base Height Area of the Parallelogram a. \[20cm\] \[246c = height\] \[1.05cm = height\] Therefore, the height of the parallelogram is \[1.05cm\] . So, the final table is S.No. Base Height Area of Triangle a. \[15cm\] \[87c = height\] \[15.5cm = height\] 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 \[12cm\] ...

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.2 Question 1. Find the value of the unknown exterior angle x in each of the following diagrams: Solution: Question 2. Find the value of the unknown interior angle x in each of the following diagrams: Solution: Question 3. Find the value of x in each of the following diagrams: Solution: Question 4. Find the value of unknown x in each of the following: Solution: Question 5. Find the values of x and y in each of the following diagrams: Solution: Question 6. Find the values of x and y in each of the following diagrams: Solution: Question 7. In the adjoining figure, find the size of each lettered angle. Solution: Question 8. One of the angles of a triangle measures 80° and the other two angles are equal. Find the measure of each of the equal angles. Solution: Question 9. If one angle of a triangle is 60° and the other two angles are in the ratio 2 : 3, find these angles. Solution: Question 10. If the angles of a triangle are in the ratio 1 : 2 : 3, find the angles. Classify the triangle in two different ways. Solution: Question 11. Can a triangle have three angles whose measures are (i) 65°, 74°, 39°? (ii) \(\frac \) right angle, 1 right angle, 60°? Solution: Categories Post navigation

Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take `pi = 22/7`) Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find 1)the area covered by the roads. 2)the cost of constructing the roads at the rate of Rs 110 per m 2. Shaalaa.com has the CBSE Mathematics Class 7 Maths CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and provide excellent self-help guidance for students. Concepts covered in Using NCERT Class 7 Maths solutions Perimeter and Area exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE Class 7 Maths students prefer NCERT Textbook Solutions to score more in exams. Get the free...

This article will help to solve the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2 and also, check out the important formula: NCERT Solutions Class 7th Math Chapter 11Details Textbook NCERT Class 7 th Subject Mathematics Chapter 11 th Chapter Perimeter and Area Exercise 11.2 Category Class 7th Maths Medium English Formula Used for Maths Ncert Solution Class 7 Chapter 11 Area of square a 2 Area of a rectangle l x b Area of a triangle ½ x b x h Area of a circle πr 2 NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2 Ex 11.2 Class 7 Maths Question 1: Find the area of each of the following parallelograms: (a) Solution:Given that, Height of parallelogram = 4 cm Base of parallelogram = 7 cm Then, Area of parallelogram = base × height = 7 × 4 cm 2 = 28 cm 2 (b) Solution:Given that, Height of parallelogram = 3 cm Base of parallelogram = 5 cm Then, Area of parallelogram = base × height = 5 × 3 = 15 cm 2 (c) Solution:Given That, Height of parallelogram = 3.5 cm Base of parallelogram = 2.5 cm Then, Area of parallelogram = base × height = 2.5 × 3.5 = 8.75 cm 2 (d) Solution:Given That, Height of parallelogram = 4.8 cm Base of parallelogram = 5 cm Then, Area of parallelogram = base × height = 5 × 4.8 = 24 cm 2 Solution:Given that, Height of parallelogram = 4.4 cm Base of parallelogram = 2 cm Then, Area of parallelogram = base × height = 2 × 4.4 = 8.8 cm 2 Ex 11.2 Class 7 Maths Question 2: Find the area of each of the following tr...