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Exercise 2.4 · Q3

Q.A rope by which a mare is tied at a centre of the circular field is decreased from 23m to 11 m.

a) What is the decrease in area more will be able to graze now?
b) Find the percentage decrease in area grazed? due to decreased length of rope.
Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
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✓ Free question

A tethered mare's rope is shortened from 23 m to 11 m; the grazing area (a circle of radius = rope length) decreases by ≈1282.29 m², a ≈77.13% drop.

The area a tethered animal can graze is a circle with radius equal to the rope length: A=πr2A=\pi r^2. Percentage decrease =decreaseoriginal area×100=\dfrac{\text{decrease}}{\text{original area}}\times100.

  1. Original rope length (radius) r1=23r_1=23 m ⇒\Rightarrow Area1=π(23)2=529π m2_1=\pi(23)^2=529\pi\text{ m}^2.
  2. New rope length r2=11r_2=11 m ⇒\Rightarrow Area2=π(11)2=121π m2_2=\pi(11)^2=121\pi\text{ m}^2.
  3. (a) Decrease in area == Area1−_1-Area2=π(529−121)=408π_2=\pi(529-121)=408\pi.
  4. Using π=227\pi=\dfrac{22}{7}: 408π=408×227=89767≈1282.29 m2408\pi=\dfrac{408\times22}{7}=\dfrac{8976}{7}\approx1282.29\text{ m}^2.
  5. (b) Percentage decrease =408π529π×100=408529×100≈77.13%=\dfrac{408\pi}{529\pi}\times100=\dfrac{408}{529}\times100\approx77.13\%.
  6. Self-check: remaining area fraction =121529×100≈22.87%=\dfrac{121}{529}\times100\approx22.87\%; 22.87%+77.13%=100%22.87\%+77.13\%=100\% ✓.
✓Final answer

a) The grazing area decreases by ≈\approx 1282.29 m²; b) this is a decrease of ≈\approx 77.13%.

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