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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 11.1, 8 A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find the area of that part of the field in which the horse can graze.(Use π = 3.14) Let ABCD be square field And, length of rope = 5 m ∴ r = 5 m We need to find area of field which horse can graze, i.e. Area of sector QBP Since In square all angles are 90°. Hence, ∠ QBP = 90° Now, Area of sector QBP = 𝜃/360×πr2 = 𝟗𝟎/𝟑𝟔𝟎 × 𝟑.𝟏𝟒 × 𝟓𝟐 = 1/4 × 3.14 × 5 × 5 = (3.14 × 25)/4 = 19.625 Hence, Area of field which horse can graze = Area of sector QBP = 19.625 m2 Ex 11.1, 8 A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14) Length of the rope is increased to 10 m. Area grazed by horse now = Area of sector HBG Now, Area of sector HBG = 𝜃/360×𝜋𝑟2 = 𝟗𝟎/𝟑𝟔𝟎× 𝟑.𝟏𝟒 × (10)2 = 1/4×3.14×10×10 = 1/4 × 314 = 78.5 m2 Thus, Area grazed by horse now = 78.5 m2 Increase in grazing area = Area grazed by horse now – Area grazed previously = (Area of sector HGB – Area of sector QBP) = (78.5 – 19.625) = 58.875 m2 ∴ Increase in grazing area is 58.875 m2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.