Solution : (i) In figure (i) Since DE || BC, therefore \(AD\over DB\) = \(AE\over EC\) \(\implies\) \(1.5\over 3\) = \(1\over EC\) \(\implies\) EC = 2 cm. (ii) In figure (ii), Since DE || BC, therefore \(AD\over DB\) = \(AE\over EC\) \(\implies\) \(AD\over 7.2\) = \(1.8\over 5.4\) \(\implies\) AD = 2.4 cm.
Solution : In the given figures since sides are proportional but corresponding angles are not equal. Hence, we can say that these quadrilateral are not similar.
Solution : (i) Pair of similar figures examples are : (a) any two circles. (b) any two squares. (ii) Pair of non-similar figures examples are : (a) scalene triangle and equilateral triangle. (b) equilateral triangle and right angled triangle.
Question : Fill in the blanks using the correct word given in the brackets : (i) All circles are ………….. . (congruent, similar) (ii) All squares are ……….. . (similar, congruent). (iii) All ……………. triangles are similar. (isosceles, equilateral) Solution : (i) Similar (ii) similar (iii) equilateral
Solution : We know that the sum of opposite angles of cyclic quadrilateral is 180 degrees. Angles A and C, Angles B and D form pairs of opposite angles in the given cyclic quadrilateral ABCD. \(\angle\)A + \(\angle\)C = 180 and \(\angle\)B + \(\angle\)D = 180 \(\implies\) (4y + 20) + 4x = 180 and …
Question : Solve the following pair of linear equations : (i) px + qy = p – q and qx – py = p + q (ii) ax + by = c and bx + ay = 1 + c (iii) \(x\over a\) – \(y\over b\) = 0 …
Solution : The given linear equations are 5x – y = 5 ………(1) 3x – y = 3 ………(2) From equation (1), y = 5x – 5 When x = 1, y = 5 – 5 = 0 When x = 2, y = 10 – …
Solution : Given, \(\angle\)C = 2(\(\angle\)A + \(\angle\)B) …….(1) Adding 2\(\angle\)C on both sides in equation (1), we get \(\angle\)C + 2\(\angle\)C = 2(\(\angle\)A + \(\angle\)B) + 2\(\angle\)C \(\implies\) 3\(\angle\)C = 2(\(\angle\)A + \(\angle\)B + \(\angle\)C) Since, \(\angle\)A + \(\angle\)B + \(\angle\)C = 180 degrees \(\implies\) \(\angle\)C = \(2\over 3\) \(\times\) …
Solution : Let students in each row be x. And Let number of rows be y. then, total number of students = xy Case 1 : When 3 students are extra in a row, then number of rows becomes (y – 1) xy = (x + 3)(y – 1) \(\implies\) x – 3y + 3 …
Solution : Let x km/hr be the original speed of the train and y hrs be the time taken by the train to complete the journey. Then, Distance covered = xy km Case 1 : When speed = (x + 10) km/hr time taken = (y – 2) hr Distance = (x + 10) (y …
