Ex 11.1, 8 - Chapter 11 Class 10 Areas related to Circles

Last updated at April 16, 2024 by Teachoo

Transcript

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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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