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 Fig. 12.11). Find (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m ? (Use π = 3.14)

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Fig : 12.11

Solution:

⇨ Given

r = 5 m

𝜃 = 90°

side = 15 m

(i) Area of the field in which the horse can graze

$(i) Area of the field in which the horse can graze$

$=\frac{90}{360}\times 3.14\times \left ( 5 \right )^{2}$

$=\frac{3.14\times 25}{4}$

$=\frac{78.5}{4}$

$=19.625 m^{2}$

⇨ Area of field with 10 m robe

$⇨ Area of field with 10 m robe$

$=\frac{90}{360}\times 3.14 \left ( 10 \right )^{2}$

$=\frac{3.14\times 100}{4}$

$=\frac{314}{4}$

$=78.5 m^{2}$

(ii) Increase in the grazing area

= Area of field with 10 m robe - Area of field with 5 m robe

= (78.5 - 19.625) $m^{2}$

= 58.875 $m^{2}$

Exercise 12.2