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CAT 2018 Slot 1 — QA Question 17

Basics of CirclesEasy

In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

Answer & solution

Correct answer: (π/3√3) 1/2

  • A

    (π/6)1/2

  • (π/3√3)1/2

  • C

    (π/4√3)1/2

  • D

    (π/4)1/2

Solution

Easy

The region RR is a 60∘60^\circ sector. Triangle OCDOCD has OC=ODOC=OD and the included angle is 60∘60^\circ, which forces it to be equilateral. Equate its area to half the sector's area and solve for OCOC.

O C D A B 60°
1

Triangle OCDOCD is equilateral. With OC=OD=xOC=OD=x and the included angle ∠COD=60∘\angle COD = 60^\circ, the base angles are equal and also 60∘60^\circ, so all sides equal xx:

Area(△OCD)=34 x2\begin{aligned} &\text{Area}(\triangle OCD) = \frac{\sqrt3}{4}\,x^2 \end{aligned}
2

Area of region RR (the sector). A 60∘60^\circ slice of a circle of radius 11:

Area(R)=60360 π(1)2=π6\begin{aligned} &\text{Area}(R) = \frac{60}{360}\,\pi(1)^2 = \frac{\pi}{6} \end{aligned}
3

Apply the half-area condition. Set the triangle's area to half of RR (from steps 1 and 2):

34 x2=12⋅π6⇒ 34 x2=π12⇒ x2=π12⋅43=π33\begin{aligned} &\frac{\sqrt3}{4}\,x^2 = \frac12\cdot\frac{\pi}{6}\\ &\Rightarrow\ \frac{\sqrt3}{4}\,x^2 = \frac{\pi}{12}\\ &\Rightarrow\ x^2 = \frac{\pi}{12}\cdot\frac{4}{\sqrt3} = \frac{\pi}{3\sqrt3} \end{aligned}
4

Take the square root.

x=(π33)1/2\begin{aligned} &x = \left(\frac{\pi}{3\sqrt3}\right)^{1/2} \end{aligned}
OC=(π33)1/2 cmOC = \left(\dfrac{\pi}{3\sqrt3}\right)^{1/2}\ \text{cm}

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CAT 2018 Slot 1 QA Q17: In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bound — Solution | TheCATExam