    1. Chapter 12 Class 10 Areas related to Circles
2. Serial order wise
3. Ex 12.2

Transcript

Ex 12.2, 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 Hence (r) = 5 m We need to find area of field which horse can graze, i.e. area of sector QBP Note:- In square all all angles are 90°. Hence, ∠ QBP = 90° 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 12.2, 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 Area of sector HGB = 𝜃/360×𝜋𝑟2 = 90/360×𝜋 (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 Hence, increase in grazing area = 58.875 m2

Ex 12.2 