With vertices A, B and C of ΔABC as centres, arcs are drawn with radii 14 cm and the three portions of the triangle so obtained are removed. Find the total area removed from the triangle.
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 April 16, 2024 by Teachoo
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