Let O be the intersection point of the diagonals AC and BD of parallelogram ABCD. Since the diagonals of a parallelogram bisect each other, OA = OC and OB = OD. Given that DP = BQ, we have OP = OD - DP = OB - BQ = OQ. The diagonals AC and PQ of quadrilateral APCQ bisect each other at O, which provesRead more
Let O be the intersection point of the diagonals AC and BD of parallelogram ABCD. Since the diagonals of a parallelogram bisect each other, OA = OC and OB = OD. Given that DP = BQ, we have OP = OD – DP = OB – BQ = OQ. The diagonals AC and PQ of quadrilateral APCQ bisect each other at O, which proves APCQ is a parallelogram.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
The crossing forms two triangles, △EAD and △EBC, meeting at intersection point E. The sum of all angles in both triangles is 360°, giving (∠A + ∠D + ∠AED) + (∠B + ∠C + ∠BEC) = 360°. Because ∠AED = ∠BEC > 0°, the sum ∠A + ∠B + ∠C + ∠D = 360° - 2∠AED < 360°. Yes, choosing lines intersecting withRead more
The crossing forms two triangles, △EAD and △EBC, meeting at intersection point E. The sum of all angles in both triangles is 360°, giving (∠A + ∠D + ∠AED) + (∠B + ∠C + ∠BEC) = 360°. Because ∠AED = ∠BEC > 0°, the sum ∠A + ∠B + ∠C + ∠D = 360° – 2∠AED < 360°. Yes, choosing lines intersecting with ∠AED = 179° produces an angle sum of exactly 2°.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
Count the horizontal ruled lines intersecting the segment: its endpoints lie on the first and fifth lines, spanning four equally spaced intervals. The third line lies precisely halfway between them. By the Converse of the Midpoint Theorem and the equal intercept theorem, the intersection of the segmRead more
Count the horizontal ruled lines intersecting the segment: its endpoints lie on the first and fifth lines, spanning four equally spaced intervals. The third line lies precisely halfway between them. By the Converse of the Midpoint Theorem and the equal intercept theorem, the intersection of the segment with this third horizontal line divides the segment into two equal halves, marking its exact midpoint.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
Rotate a copy of the triangle by 180° and join it with the original along an equal side to form a parallelogram. As established, identical parallelograms can tile the plane seamlessly without overlaps or gaps. By decomposing every parallelogram in that grid into the two original congruent trianglesRead more
Rotate a copy of the triangle by 180° and join it with the original along an equal side to form a parallelogram. As established, identical parallelograms can tile the plane seamlessly without overlaps or gaps. By decomposing every parallelogram in that grid into the two original congruent triangles along their diagonals, we obtain a complete tiling of the plane using copies of the triangle.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
Let x denote adult tickets and y denote children's tickets. Because 200 people attended the event, the sum of tickets sold is expressed as x + y = 200. Adults paid ₹150 and children paid ₹100, generating a total revenue of ₹25,000, which gives 150x + 100y = 25000 (or 3x + 2y = 500). These two equatiRead more
Let x denote adult tickets and y denote children’s tickets. Because 200 people attended the event, the sum of tickets sold is expressed as x + y = 200. Adults paid ₹150 and children paid ₹100, generating a total revenue of ₹25,000, which gives 150x + 100y = 25000 (or 3x + 2y = 500). These two equations represent the complete situation.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 12.34). Show that APCQ is a parallelogram.
Let O be the intersection point of the diagonals AC and BD of parallelogram ABCD. Since the diagonals of a parallelogram bisect each other, OA = OC and OB = OD. Given that DP = BQ, we have OP = OD - DP = OB - BQ = OQ. The diagonals AC and PQ of quadrilateral APCQ bisect each other at O, which provesRead more
Let O be the intersection point of the diagonals AC and BD of parallelogram ABCD. Since the diagonals of a parallelogram bisect each other, OA = OC and OB = OD. Given that DP = BQ, we have OP = OD – DP = OB – BQ = OQ. The diagonals AC and PQ of quadrilateral APCQ bisect each other at O, which proves APCQ is 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 lessYou know that the sum of angles of a quadrilateral is 360°, even for a non-convex quadrilateral. (Recall the proof.) Now consider a self-intersecting quadrilateral ABCD, where AB and CD intersect at point E. Show that ∠A + ∠B + ∠C + ∠D < 360°. Can you construct ABCD such that ∠A + ∠B + ∠C + ∠D = 2°?
The crossing forms two triangles, △EAD and △EBC, meeting at intersection point E. The sum of all angles in both triangles is 360°, giving (∠A + ∠D + ∠AED) + (∠B + ∠C + ∠BEC) = 360°. Because ∠AED = ∠BEC > 0°, the sum ∠A + ∠B + ∠C + ∠D = 360° - 2∠AED < 360°. Yes, choosing lines intersecting withRead more
The crossing forms two triangles, △EAD and △EBC, meeting at intersection point E. The sum of all angles in both triangles is 360°, giving (∠A + ∠D + ∠AED) + (∠B + ∠C + ∠BEC) = 360°. Because ∠AED = ∠BEC > 0°, the sum ∠A + ∠B + ∠C + ∠D = 360° – 2∠AED < 360°. Yes, choosing lines intersecting with ∠AED = 179° produces an angle sum of exactly 2°.
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 lessMark the midpoint of the line drawn on the paper (see Fig. 12.33), given that the horizontal lines are equally spaced. Justify your answer.
Count the horizontal ruled lines intersecting the segment: its endpoints lie on the first and fifth lines, spanning four equally spaced intervals. The third line lies precisely halfway between them. By the Converse of the Midpoint Theorem and the equal intercept theorem, the intersection of the segmRead more
Count the horizontal ruled lines intersecting the segment: its endpoints lie on the first and fifth lines, spanning four equally spaced intervals. The third line lies precisely halfway between them. By the Converse of the Midpoint Theorem and the equal intercept theorem, the intersection of the segment with this third horizontal line divides the segment into two equal halves, marking its exact midpoint.
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 lessUsing a fact about parallelograms, show how to tile the plane using any given triangle. (Hint: Can you use the parallelogram tiling in the introduction?)
Rotate a copy of the triangle by 180° and join it with the original along an equal side to form a parallelogram. As established, identical parallelograms can tile the plane seamlessly without overlaps or gaps. By decomposing every parallelogram in that grid into the two original congruent trianglesRead more
Rotate a copy of the triangle by 180° and join it with the original along an equal side to form a parallelogram. As established, identical parallelograms can tile the plane seamlessly without overlaps or gaps. By decomposing every parallelogram in that grid into the two original congruent triangles along their diagonals, we obtain a complete tiling of the plane using copies of the triangle.
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 lessIn a sports meet, tickets for adults cost ₹150 each and tickets for children cost ₹100 each. A total of 200 people attended the meet and the total amount collected from ticket sales was ₹25,000. (Let the number of adult tickets sold be x and the number of children’s tickets sold be y.) Frame a pair of linear equations in x and y to represent this situation.
Let x denote adult tickets and y denote children's tickets. Because 200 people attended the event, the sum of tickets sold is expressed as x + y = 200. Adults paid ₹150 and children paid ₹100, generating a total revenue of ₹25,000, which gives 150x + 100y = 25000 (or 3x + 2y = 500). These two equatiRead more
Let x denote adult tickets and y denote children’s tickets. Because 200 people attended the event, the sum of tickets sold is expressed as x + y = 200. Adults paid ₹150 and children paid ₹100, generating a total revenue of ₹25,000, which gives 150x + 100y = 25000 (or 3x + 2y = 500). These two equations represent the complete situation.
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