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Virat
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Asked: 2026-10-03T06:00:36+00:00 In:

You 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°?

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In triangles EAD and EBC, vertical angles at E are equal, so the four vertex angles sum to 360° – 2∠AED < 360°. Yes, make angle E nearly 179° so the remaining sum equals 2°.

Cbse released Class 9 Ganita Manjari Part 2 book
Class 9 Chapter 12 Quadrilaterals (Ganita Manjari 2) Solutions

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    1. 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/

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