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A horse, a cow and a goat are tied, each by ropes of length 14m - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard

part 2 - Question 24 (A) - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 24 (A) - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10 part 4 - Question 24 (A) - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10

 

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Question 24 (A) A horse, a cow and a goat are tied, each by ropes of length 14m, at the corners A, B and C respectively, of a grassy triangular field ABC with sides of lengths 35 m, 40 m and 50 m. Find the area of grass field that can be grazed by them. Here, the animals graze area of sectors Area of grass grazed = Area of Sector grazed by all three animals Area of Sector grazed by all three animals We know that Area of sector = 𝜽/(𝟑𝟔𝟎°) × πr2 And radius = 14 m Now, 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, Area of sector grazed by all 3 animals = (∠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 = 14 m = 1/2 × 22/7 × (14)2 = 11/7 × 14 × 14 = 11 × 2 × 14 = 308 m2

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