Since (2, -1) satisfies both equations, substitute x = 2 and y = -1 into each. For 2mx + 3y = 7, we get 2m(2) + 3(-1) = 7, leading to 4m - 3 = 7, which simplifies to m = 2.5 or 5/2. For 4x + ny = -10, we get 4(2) + n(-1) = -10, simplifying to 8 - n = -10, which gives n = 18. For more NCERT SoRead more
Since (2, -1) satisfies both equations, substitute x = 2 and y = -1 into each. For 2mx + 3y = 7, we get 2m(2) + 3(-1) = 7, leading to 4m – 3 = 7, which simplifies to m = 2.5 or 5/2. For 4x + ny = -10, we get 4(2) + n(-1) = -10, simplifying to 8 – n = -10, which gives n = 18.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
i) True: Consecutive interior angles are supplementary, so having one 90° angle forces all angles to equal 90°, forming a rectangle. (ii) False: Diagonals of every rhombus already intersect perpendicularly; it requires equal diagonals or one right angle to become a square. (iii) True: In any paralleRead more
i) True: Consecutive interior angles are supplementary, so having one 90° angle forces all angles to equal 90°, forming a rectangle.
(ii) False: Diagonals of every rhombus already intersect perpendicularly; it requires equal diagonals or one right angle to become a square.
(iii) True: In any parallelogram with equal diagonals, opposite triangles are congruent by SSS, proving the internal angles are right angles.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
Let diagonal AC bisect ∠A, so ∠DAC = ∠BAC. By alternate interior angles, ∠BAC = ∠DCA and ∠DAC = ∠BCA. Hence, ∠DCA = ∠BCA, proving AC bisects ∠C. Furthermore, in △ABC, ∠BAC = ∠BCA implies AB = BC. Since adjacent sides of this parallelogram are equal, all four sides are equal, confirming ABCD is a rhoRead more
Let diagonal AC bisect ∠A, so ∠DAC = ∠BAC. By alternate interior angles, ∠BAC = ∠DCA and ∠DAC = ∠BCA. Hence, ∠DCA = ∠BCA, proving AC bisects ∠C. Furthermore, in △ABC, ∠BAC = ∠BCA implies AB = BC. Since adjacent sides of this parallelogram are equal, all four sides are equal, confirming ABCD is a rhombus.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
(i) Yes: Diagonals bisecting vertex angles force all opposite sides parallel and adjacent sides equal, creating a rhombus. (ii) Yes: Perpendicular bisecting diagonals prove all four sides are congruent through SAS triangle congruences. (iii) No: Non-rectangular shapes like isosceles trapeziums haveRead more
(i) Yes: Diagonals bisecting vertex angles force all opposite sides parallel and adjacent sides equal, creating a rhombus.
(ii) Yes: Perpendicular bisecting diagonals prove all four sides are congruent through SAS triangle congruences.
(iii) No: Non-rectangular shapes like isosceles trapeziums have equal diagonals. Extra condition: Add that the diagonals must bisect each other or the quadrilateral must already be a parallelogram.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
Adjacent angles of parallelogram ABCD are supplementary, summing to 180°. In triangles formed by adjacent bisectors, half-angles sum to 90°, forcing each intersection angle to be 90°. As all four interior angles equal 90°, the figure PQRS is a rectangle. We assume AB ≠ BC because if AB = BC (a rhombRead more
Adjacent angles of parallelogram ABCD are supplementary, summing to 180°. In triangles formed by adjacent bisectors, half-angles sum to 90°, forcing each intersection angle to be 90°. As all four interior angles equal 90°, the figure PQRS is a rectangle. We assume AB ≠ BC because if AB = BC (a rhombus), the angle bisectors are the diagonals, which concur at one point rather than forming a quadrilateral.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
In the equations given below, m and n are unknown constants: 2mx + 3y = 7; 4x + ny = -10. If (2, -1) is the solution of both equations, find the values of m and n.
Since (2, -1) satisfies both equations, substitute x = 2 and y = -1 into each. For 2mx + 3y = 7, we get 2m(2) + 3(-1) = 7, leading to 4m - 3 = 7, which simplifies to m = 2.5 or 5/2. For 4x + ny = -10, we get 4(2) + n(-1) = -10, simplifying to 8 - n = -10, which gives n = 18. For more NCERT SoRead more
Since (2, -1) satisfies both equations, substitute x = 2 and y = -1 into each. For 2mx + 3y = 7, we get 2m(2) + 3(-1) = 7, leading to 4m – 3 = 7, which simplifies to m = 2.5 or 5/2. For 4x + ny = -10, we get 4(2) + n(-1) = -10, simplifying to 8 – n = -10, which gives n = 18.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See less(i) A parallelogram with a right angle is a rectangle. (ii) A rhombus with perpendicular diagonals is a square. (iii) If the diagonals of a parallelogram are equal, then it is a rectangle.
i) True: Consecutive interior angles are supplementary, so having one 90° angle forces all angles to equal 90°, forming a rectangle. (ii) False: Diagonals of every rhombus already intersect perpendicularly; it requires equal diagonals or one right angle to become a square. (iii) True: In any paralleRead more
i) True: Consecutive interior angles are supplementary, so having one 90° angle forces all angles to equal 90°, forming a rectangle.
(ii) False: Diagonals of every rhombus already intersect perpendicularly; it requires equal diagonals or one right angle to become a square.
(iii) True: In any parallelogram with equal diagonals, opposite triangles are congruent by SSS, proving the internal angles are right angles.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessThe diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.
Let diagonal AC bisect ∠A, so ∠DAC = ∠BAC. By alternate interior angles, ∠BAC = ∠DCA and ∠DAC = ∠BCA. Hence, ∠DCA = ∠BCA, proving AC bisects ∠C. Furthermore, in △ABC, ∠BAC = ∠BCA implies AB = BC. Since adjacent sides of this parallelogram are equal, all four sides are equal, confirming ABCD is a rhoRead more
Let diagonal AC bisect ∠A, so ∠DAC = ∠BAC. By alternate interior angles, ∠BAC = ∠DCA and ∠DAC = ∠BCA. Hence, ∠DCA = ∠BCA, proving AC bisects ∠C. Furthermore, in △ABC, ∠BAC = ∠BCA implies AB = BC. Since adjacent sides of this parallelogram are equal, all four sides are equal, confirming ABCD is a rhombus.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessThe following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes? (i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus? (ii) If the diagonals of a quadrilateral ABCD bisect each other at right angles, must ABCD be a rhombus? (iii) If the diagonals of a quadrilateral ABCD are of equal length, must ABCD be a rectangle?
(i) Yes: Diagonals bisecting vertex angles force all opposite sides parallel and adjacent sides equal, creating a rhombus. (ii) Yes: Perpendicular bisecting diagonals prove all four sides are congruent through SAS triangle congruences. (iii) No: Non-rectangular shapes like isosceles trapeziums haveRead more
(i) Yes: Diagonals bisecting vertex angles force all opposite sides parallel and adjacent sides equal, creating a rhombus.
(ii) Yes: Perpendicular bisecting diagonals prove all four sides are congruent through SAS triangle congruences.
(iii) No: Non-rectangular shapes like isosceles trapeziums have equal diagonals. Extra condition: Add that the diagonals must bisect each other or the quadrilateral must already be a parallelogram.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessLet ABCD be a parallelogram with AB ≠ BC. Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle (see Fig. 12.11). Why did we assume AB ≠ BC?
Adjacent angles of parallelogram ABCD are supplementary, summing to 180°. In triangles formed by adjacent bisectors, half-angles sum to 90°, forcing each intersection angle to be 90°. As all four interior angles equal 90°, the figure PQRS is a rectangle. We assume AB ≠ BC because if AB = BC (a rhombRead more
Adjacent angles of parallelogram ABCD are supplementary, summing to 180°. In triangles formed by adjacent bisectors, half-angles sum to 90°, forcing each intersection angle to be 90°. As all four interior angles equal 90°, the figure PQRS is a rectangle. We assume AB ≠ BC because if AB = BC (a rhombus), the angle bisectors are the diagonals, which concur at one point rather than forming a quadrilateral.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See less