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Tethered Cow Grazing Area

CAT 1998 Slot 1 · Quantitative Ability · Hard · Geometry

This is a hard Quantitative Ability question set from the CAT 1998 Slot 1 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Passage / data set

A cow is tethered at point A by a rope. Neither the rope nor the cow is allowed to enter ΔABC\Delta ABC.

∠BAC=30∘\angle BAC = 30^\circ AB=AC=10 mAB = AC = 10\text{ m}

Question 1 of 2

What is the area that can be grazed by the cow if the length of the rope is 8 m8\text{ m}?

  1. A.

    13413π sq. m134\frac{1}{3}\pi\text{ sq. m}

  2. B.

    121π sq. m121\pi\text{ sq. m}

  3. C.

    132π sq. m132\pi\text{ sq. m}

  4. D.

    1763π sq. m\frac{176}{3}\pi\text{ sq. m}

Answer

D

Explanation

The cow can graze a major sector with angle 360∘−30∘=330∘360^\circ - 30^\circ = 330^\circ. Area =330∘360∘×π×82=1112×64π=1763π sq. m= \frac{330^\circ}{360^\circ} \times \pi \times 8^2 = \frac{11}{12} \times 64\pi = \frac{176}{3}\pi\text{ sq. m}.

Question 2 of 2

What is the area that can be grazed by the cow if the length of the rope is 12 m12\text{ m}?

  1. A.

    13316π sq. m133\frac{1}{6}\pi\text{ sq. m}

  2. B.

    121π sq. m121\pi\text{ sq. m}

  3. C.

    132π sq. m132\pi\text{ sq. m}

  4. D.

    1763π sq. m\frac{176}{3}\pi\text{ sq. m}

Answer

A

Explanation

Main sector area =330360×π×122=132π= \frac{330}{360} \times \pi \times 12^2 = 132\pi. When rope reaches B or C, remaining length =12−10=2 m= 12 - 10 = 2\text{ m}. Additional grazing at vertices B and C adds extra sector areas, resulting in slightly more than 132π132\pi, specifically 13316π sq. m133\frac{1}{6}\pi\text{ sq. m}.

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