CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard
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CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard
Last updated at July 18, 2026 by Teachoo
Transcript
Total area removed from the triangle = Area of sector at point A + Area of sector at point B + Area of sector at point C Area of Sector We know that Area of sector = 𝜃/(360°) × πr2 and radius = 14 cm Thus, Area of Sector at point A = (∠𝐴)/(360°) × πr2 Area of Sector at point B = (∠𝐵)/(360°) × πr2 Area of Sector at point C = (∠𝐶)/(360°) × πr2 Therefore, Total area removed from the triangle = (∠𝐴)/(360°) × πr2 + (∠𝐵)/(360°) × πr2 + (∠𝐶)/(360°) × πr2 = 𝟏/(𝟑𝟔𝟎°) × πr2 (∠ A + ∠ B + ∠ C) Since sum of angles of a triangle = 180° = 1/(360°) × πr2 × 180° = 𝟏/𝟐 × πr2 Putting r = 14cm = 1/2 × 22/7 × (14)2 = 11/7 × 14 × 14 = 11 × 28 = 308 〖𝒄𝒎〗^𝟐
