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Show that the sum of angles of a non-planar quadrilateral is always less than 360°. Can you find a non-planar quadrilateral ABCD for which ∠A + ∠B + ∠C + ∠D = 2°? (Hint: Think of a diagonal, say AC, as a hinge around which triangles ABC and ADC can rotate.) What happens to each angle of ABCD as you do this rotation?
Folding along diagonal AC treats AC as a spatial hinge between triangles △ABC and △ADC. By 3D angle inequalities, the spatial angle is strictly less than the sum of its planar face angles: ∠A < ∠BAC + ∠CAD and ∠C < ∠BCA + ∠ACD. Since ∠B and ∠D remain unchanged, the four angles sum to strictlyRead more
Folding along diagonal AC treats AC as a spatial hinge between triangles △ABC and △ADC. By 3D angle inequalities, the spatial angle is strictly less than the sum of its planar face angles: ∠A < ∠BAC + ∠CAD and ∠C < ∠BCA + ∠ACD. Since ∠B and ∠D remain unchanged, the four angles sum to strictly less than 360°. Yes, folding two extremely narrow triangles nearly flat yields a 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 lessLet us see a third method to tile the plane using a 4-gon. Focus on only two coloured copies of SOME sharing a vertex. What do you see? It appears that each 4-gon is just a shifted copy of the other. Let us see exactly how. Draw two copies of SOME as shown in Fig. 12.42 so that the diagonals EO and E’O’ are collinear with O = E’. Now place a cutout of one copy on top of SOME with diagonal EO drawn on it. Slide this cutout so that segment EO moves along EO’ until EO matches E’O’. Verify that your cutout exactly matches S’O’M’E’. (i) Prove the exact match of SOME with S’O’M’E’ using four parallelograms. (ii) Follow the described procedure along each diagonal of the starting coloured 4-gon in both directions to get 4 new 4-gons. Using these 4 copies, repeat the procedure 4 more times to place 4 new copies, resulting in a 3 by 3 grid of 9 copies exactly like the 9 coloured copies in Fig. 12.42. These 9 copies meet each other only at vertices and once again the blank spaces among them also form 4-gons congruent to SOME!
(i) Translating the quadrilateral by the vector EO shifts each vertex by the same distance and direction. This translation creates four parallelograms connecting original vertices to shifted counterparts: E E' O' O, E E' S' S and E E' M' M. Because translation is a rigid motion preserving all side lRead more
(i) Translating the quadrilateral by the vector EO shifts each vertex by the same distance and direction. This translation creates four parallelograms connecting original vertices to shifted counterparts: E E’ O’ O, E E’ S’ S and E E’ M’ M. Because translation is a rigid motion preserving all side lengths and angles, S’O’M’E’ is identically congruent to SOME. (ii) Repeating this translation along both diagonal directions forms a regular grid of nine quadrilaterals touching at vertices, leaving interior voids that perfectly fit inverted copies of SOME.
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 lessIs there a 4-gon with given side lengths? (i) Recall the following fact about triangles and check it by construction. For given positive numbers a, b, c, is there a triangle whose sides have these lengths? The answer is Yes exactly when the sum of any two numbers is greater than the third. If we arrange the numbers in increasing order (suppose a ≤ b ≤ c), then this amounts to requiring a + b > c. (Hint: Start by drawing a segment of length c.) (ii) Suppose a 4-gon has 2, 5, 11 as three side lengths. Can the length of the fourth side be 100? Can it be 10? Can it be 1? What are the possible lengths of the fourth side? (iii) For given positive numbers a, b, c, d, how will you decide if there is a 4-gon whose sides have these lengths?
(ii) A polygon can close only when its longest side is strictly less than the sum of all remaining sides. For side lengths 2, 5, 11 and fourth side x, we must have x < 2 + 5 + 11 = 18 and 11 < 2 + 5 + x, which gives 4 < x < 18. Hence, 100 is impossible, 1 is impossible and 10 is possibleRead more
(ii) A polygon can close only when its longest side is strictly less than the sum of all remaining sides. For side lengths 2, 5, 11 and fourth side x, we must have x < 2 + 5 + 11 = 18 and 11 < 2 + 5 + x, which gives 4 < x < 18. Hence, 100 is impossible, 1 is impossible and 10 is possible. (iii) A 4-gon exists if and only if each side is strictly smaller than the sum of the other three sides.
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 lessCounting diagonals of a polygon. (i) How should we define a diagonal of an n-gon? How many diagonals does an n-gon have? Make a table for small values of n. A 3-gon has no diagonals. A 4-gon has 2. How many diagonals does a 5-gon have? A 6-gon? Try to find a pattern and guess the answers for n = 7 and n = 8. Check your guesses by systematic counting. (ii) Can you guess a formula for the number of diagonals? How many diagonals get added when we increase the number of sides by 1? Can you now justify why the formula you guessed is true for all n?
(i) A diagonal is a straight line segment joining any two non-consecutive vertices. Counts: 3-gon has 0; 4-gon has 2; 5-gon has 5; 6-gon has 9; 7-gon has 14; 8-gon has 20 diagonals. (ii) Adding one vertex introduces n - 1 new diagonals. Formula: n(n - 3)/2. Justification: Each of the n vertices canRead more
(i) A diagonal is a straight line segment joining any two non-consecutive vertices. Counts: 3-gon has 0; 4-gon has 2; 5-gon has 5; 6-gon has 9; 7-gon has 14; 8-gon has 20 diagonals.
(ii) Adding one vertex introduces n – 1 new diagonals. Formula: n(n – 3)/2. Justification: Each of the n vertices can connect to n – 3 non-adjacent vertices (excluding itself and its two neighbors). Multiplying gives n(n – 3) and dividing by 2 avoids double-counting each segment.
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 lessSum of angles of a polygon. What is the sum of angles of a (planar non-self-intersecting) n-gon? We know that the answer is 180° for n = 3 and 360° for n = 4. Find the next few values. Then find a formula in terms of n and prove it.
For n = 5, the sum of interior angles is 540°; for n = 6, it is 720°; for n = 7, it is 900°. The general formula for an n-gon is (n - 2) × 180°. Proof: Choose any single vertex and draw all possible diagonals to the other vertices. This partitions the non-self-intersecting n-gon into exactly n - 2 nRead more
For n = 5, the sum of interior angles is 540°; for n = 6, it is 720°; for n = 7, it is 900°. The general formula for an n-gon is (n – 2) × 180°. Proof: Choose any single vertex and draw all possible diagonals to the other vertices. This partitions the non-self-intersecting n-gon into exactly n – 2 non-overlapping triangles. Since the angles of each triangle add up to 180°, the total sum is (n – 2) × 180°.
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/
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