Mathematics — Standard (041) — CBSE Class 10 board question

Given: radius r = 7 cm, ∠AOB = 120°, area of △OAB = 21.2 cm²

Angle of major sector OACB = 360° – 120° = 240°

(i) Perimeter of major sector OACB:

Length of arc ACB = $\dfrac{240}{360} \times 2\pi r = \dfrac{2}{3} \times 2 \times \dfrac{22}{7} \times 7 = \dfrac{88}{3}$ cm

Perimeter = arc ACB + OA + OB = $\dfrac{88}{3} + 7 + 7 = \dfrac{88}{3} + 14 = \dfrac{88 + 42}{3} = \dfrac{130}{3} \approx$ 43.33 cm

(ii) Area of shaded (minor) segment:

Area of minor sector OADB = $\dfrac{120}{360} \times \dfrac{22}{7} \times 7 \times 7 = \dfrac{1}{3} \times 154 = \dfrac{154}{3}$ cm²

Area of minor segment = Area of minor sector – Area of △OAB

$$= \frac{154}{3} - 21.2 = 51.33 - 21.2 \approx \textbf{30.13 cm}^2$$

Source: Areas Related to Circles, Chapter 11

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• The major sector angle is always (360° – given angle). For perimeter, add the two radii to the arc length — examiners commonly deduct marks if radii are omitted.
• The shaded region in the figure is the minor segment (below chord AB), so use the minor sector (120°) minus the area of △OAB.
• Use π = 22/7 as instructed in the chapter unless told otherwise.