Class 8th maths chapter 3 exercise 3.3

Mode is the value of the variable which occurs most frequently. Given below is the number of pairs of shoes of different sizes sold in a day by the owner of a shop. Size of shoe: 1, 2, 3, 4, 5, 6, 7, 8, 9 Number of pairs sold: 1, 2, 2, 3, 4, 5, 3, 7, 2 What is the modal shoe-size? Since the sale of size 8 is maximum, so the model shoe-size is 8. Bar graphs are ideal for comparing two or more values, or values over time. Data is displayed either horizontally or vertically. Single bar graphs are used to convey discrete values of an item within a category. For instance, a bar graph could display the number of males with a certain trait for specific ages. A bar graph is a pictorial representation of numerical data in the form of rectangles (or bars) of equal width and varying heights. These rectangles are drawn either vertically or horizontally, keeping equal space between them. The height (or length) of a rectangle depends upon the number it represents. Suppose some numerical data is given to us, and we have to represent it by a bar graph on a graph paper. We can draw the graph by following the steps given below: Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These lines are called the x-axis and the y-axis respectively. Step 2. Mark points at equal intervals along the x-axis. Below these points write the names of the data items whose values are to be plotted. Step 3. Choose a suitable scale. On that scale determine the heights of the bars for the...

Answer : The sum of measures of the angles of a convex quadrilateral is \(\360^\).Yes, this property holds in case if the quadrilateral is not convex. Answer : From the given table, dearly we observe that the sum of angles (interior angles) of a polygon with n sides = \((n – 2)× 180^\) Answer : (a) As we know the sum of interior angles of a quadrilateral is \(360^ \) Answer : (a)We know that the sum of interior angles of a triangle is \(180^ \) NCERT solutions for class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2 Answer : As we know that the sum of the exterior angles formed by producing the sides of a convex polygon in the same order is equal to 360°. Therefore here we have, (a)\( x + 125^ \) Answer : (i) We know that the measure of each exterior angle of a regular polygon = \(\frac\) Answer : Let the number of sides of a regular polygon be n. We have, Sum of all interior angles = \((n – 2) \times 180^=24\) So, we have the number of sides of regular polygon as 24. Answer : (a) We have, Number of sides of regular polygon=\(\frac\) But 158 does not divide 360 exactly.So, the polygon is not possible. Answer : (a) A regular polygon of 3 sides i.e, equilateral triangle has the least measure of an interior angle= \(60^\) NCERT solutions for class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3 Answer : (i) AD = BC as the opposite sides of a parallelogram are equal. (ii)∠DCB = ∠DAB as the opposite angles of a parallelogram are equal. (iii) OC = OA as ...

NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals Exercise 3.2 The NCERT Solutions for Class 8 Maths Chapter 3 – Understanding Quadrilaterals contains solutions for all exercise questions. The subject experts at BYJU’S have solved each question of NCERT exercises meticulously to help the students in solving any question from the NCERT textbook. NCERT Class 8 Exercise 3.2 is based on the measurement of the exterior angle of a polygon. Students can download the Download PDF carouselExampleControls111 Previous Next  Access other exercise solutions of Class 8 Maths Chapter 3 – Understanding Quadrilaterals Access answers of Maths NCERT Class 8 Chapter 3 – Understanding Quadrilaterals Exercise 3.2 Page Number 44 1. Find x in the following figures.  Solution: a) 125° + m = 180° ⇒ m = 180° – 125° = 55° (Linear pair) 125° + n = 180° ⇒ n = 180° – 125° = 55° (Linear pair) x = m + n (the exterior angle of a triangle is equal to the sum of 2 opposite interior 2 angles) ⇒ x = 55° + 55° = 110° b) Two interior angles are right angles = 90° 70° + m = 180° ⇒ m = 180° – 70° = 110° (Linear pair) 60° + n = 180° ⇒ n = 180° – 60° = 120° (Linear pair) The figure has five sides and is a pentagon. Thus, sum of the angles of pentagon = 540° 90° + 90° + 110° + 120° + y = 540° ⇒ 410° + y = 540° ⇒ y = 540° – 410° = 130° x + y = 180° (Linear pair) ⇒ x + 130° = 180° ⇒ x = 180° – 130° = 50° 2. Find th...

Tamilnadu State Board New Syllabus Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.3 Question 1. Expand (i) (3m + 5) 2 (ii) (5p – 1) 2 (iii) (2n – 1)(2n + 3) (iv) 4p 2– 25q 2 Answer: (i) (3m + 5) 2 Comparing (3m + 5) 2 with (a + b) 2 we have a = 3m and b = 5 (a + b) 2 = a 2 + 2ab + b 2 (3m + 5) 2 = (3m) 2 + 2(3m) (5) + 5 2 = 3 2 m 2 + 30 m + 25 = 9m 2 + 30 m + 25 (ii) (5p – 1) 2 Comparing (5p – 1) 2 with (a – b) 2 we have a = 5p and b = 1 (a – b) 2 = a 2– 2ab + b 2 (5p – 1) 2 = (5p) 2– 2(5p)(1) + 1 2 = 5 2p 2– 10 p + 1 = 25p 2– 10p + 1 (iii) (2n – 1)(2n + 3) Comparing (2n – 1) (2n + 3) with (x + a)(x + b) we have a = -1; b = 3 (x + a) (x + b) = x 2 + (a + b)x + ab (2n + (-1)) (2n + 3) = (2n) 2 + (-1 + 3)2n + (-1) (3) = 2 2 n 2 + 2(2n) – 3 = 4n 2 + 4n – 3 (iv) 4p 2– 25q 2 = (2p) 2– (5q) 2 Comparing (2p) 2– (5q) 2 with a 2– b 2 we have a = 2p and b = 5q (a 2– b 2) = (a + b)(a – b) = (2p + 5q)(2p – 5q) Question 2. Expand (i) (3 + m) 3 (ii) (2a + 5) 3 (iii) (3p + 4q) 3 (iv) (52) 3 (v) (104) 3 Answer: (i) (3 + m) 3 Cornparing (3 + m) 3 with (a + b) 3 we have a = 3 ; b = m (a + b) 3 = a 2 + 3a 2b + 3ab 2 + b 3 (3 + m) 3 = 3 3 + 3(3) 2 (m) + 3(3)m 2 + m 3 = 27 + 27m + 9m 2 + m 3 = m 3 + 9m 2 + 27m + 27 (ii) (2a + 5) 3 = Comparing (2a + 5) 3 with (a + b) 3 we have a = 2a, b = 5 (a + b) 3 = a 3 + 3a 2b + 3ab 2 + b 3 = (2a) 3 + 3(2a) 2 5 + 3 (2a) 5 2 + 5 3 = 2 3a 3 + 3(2 2a 2)5 + 6a(25) + 125 = 8a 3 + 60a 2 + 150a + 125 (iii) (3p + 4q) 3 Comparing (3p + 4q) 3 wi...

NCERT Solutions Class 8 Maths Chapter 3 Exercise 3.3 Understanding Quadrilaterals NCERT solutions for class 8 maths chapter 3 exercise 3.3 understanding quadrilaterals takes a closer look at a parallelogram as well as a trapezium and their elements. Kids can get an insight into the properties of The questions in the class 8 maths NCERT solutions chapter 3 exercise 3.3 understanding ☛ Download NCERT Solutions Class 8 Maths Chapter 3 Exercise 3.3 Exercise 3.3 Class 8 Chapter 3 Download PDF More Exercises in Class 8 Maths Chapter 3 • • • NCERT Solutions Class 8 Maths Chapter 3 Exercise 3.3 Tips • Read through the properties of a parallelogram and trapezium. • See the solved examples to get a better understanding of these theoretical concepts. • Apply the concepts to questions by yourself. • Finally use the Download Cuemath NCERT Solutions PDF for free and start learning! NCERT Video Solutions for Class 8 Maths Chapter 3 Exercise 3.3 Video Solutions for Class 8 Maths Chapter 3 Exercise 3.3