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CAT 2024 Slot 3 — QA Question 1

Basics of CirclesEasy

A circular plot of land is divided into two regions by a chord of length 10√3 meters such that the chord subtends an angle of 120° at the center. Then, the area, in square meters, of the smaller region is

Answer & solution

Correct answer: 25 4 π 3 - 3

  • 25(4π3-3)

  • B

    20(4π3+3)

  • C

    20(4π3-3)

  • D

    25(4π3+3)

Solution

Medium

The chord cuts off a circular segment. Its area equals (area of the sector swept by the central angle) minus (area of the triangle formed by the two radii and the chord). First recover the radius from the chord length and central angle.

1

Find the radius. The chord of length 10310\sqrt{3} subtends 120∘120^\circ at the centre. Drop a perpendicular from the centre to the chord; it bisects both the chord and the angle, giving a right triangle with angle 60∘60^\circ and opposite side 535\sqrt3.

sin⁡60∘=53r⇒ 32=53r(half-chord over radius)⇒ r=10\begin{aligned} &\sin 60^\circ=\frac{5\sqrt3}{r}\\ &\Rightarrow\ \frac{\sqrt3}{2}=\frac{5\sqrt3}{r}\quad\text{(half-chord over radius)}\\ &\Rightarrow\ r=10 \end{aligned}
2

Area of the sector. The smaller region sits on the 120∘120^\circ side, so use that angle.

Asector=120360 πr2⇒ Asector=13 π(10)2(from step 1, r=10)⇒ Asector=100π3\begin{aligned} &A_{\text{sector}}=\frac{120}{360}\,\pi r^2\\ &\Rightarrow\ A_{\text{sector}}=\frac13\,\pi(10)^2\quad\text{(from step 1, }r=10\text{)}\\ &\Rightarrow\ A_{\text{sector}}=\frac{100\pi}{3} \end{aligned}
3

Area of the triangle of the two radii. The triangle has two sides r=10r=10 enclosing 120∘120^\circ.

A△=12r2sin⁡120∘⇒ A△=12(100)⋅32(sin⁡120∘=32)⇒ A△=253\begin{aligned} &A_{\triangle}=\tfrac12 r^2\sin 120^\circ\\ &\Rightarrow\ A_{\triangle}=\tfrac12(100)\cdot\frac{\sqrt3}{2}\quad\text{(}\sin120^\circ=\tfrac{\sqrt3}{2}\text{)}\\ &\Rightarrow\ A_{\triangle}=25\sqrt3 \end{aligned}
4

Smaller region = segment. Subtract the triangle from the sector.

A=Asector−A△⇒ A=100π3−253[(step 2)-(step 3)]⇒ A=25 ⁣(4π3−3)\begin{aligned} &A=A_{\text{sector}}-A_{\triangle}\\ &\Rightarrow\ A=\frac{100\pi}{3}-25\sqrt3\quad\text{[(step 2)-(step 3)]}\\ &\Rightarrow\ A=25\!\left(\frac{4\pi}{3}-\sqrt3\right) \end{aligned}
120° chord = 10√3 r=10
25 ⁣(4π3−3)25\!\left(\dfrac{4\pi}{3}-\sqrt3\right)
Need a hint?

Segment area = sector area −- triangle area. Get rr from chord=2rsin⁡(θ/2)\text{chord}=2r\sin(\theta/2).

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CAT 2024 Slot 3 QA Q1 — Basics of Circles solution | TheCATExam