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ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. If the chords subtend an angle of 120º at their respective centres, find the total area of the given figure, where arcs are a part of the circle.

(a) \(\bigg[a^2+4\bigg[\frac{\pi{a}^2}{9}-\frac{a^2}{3\sqrt2}\bigg]\bigg]\)

(b) \(\bigg[a^2+4\bigg[\frac{\pi{a}^2}{9}-\frac{a^2}{4\sqrt3}\bigg]\bigg]\)

(c) [9a2 – 4π + 3 √3 a2

(d) None of these

1 Answer

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 (b) \(\bigg[a^2+4\bigg[\frac{\pi{a}^2}{9}-\frac{a^2}{4\sqrt3}\bigg]\bigg]\)

As shown in the given figures, if ‘a’ is each side of the square, then ∠DOC = 120º ⇒ ∠ODC = ∠OCD = 30º

Now in fig. (iii), \(\frac{PC}{OC}\) = sin 60º ⇒ \(\frac{\frac{a}{2}}{OC}\) = \(\frac{\sqrt3}{2}\)

⇒ OC = \(\frac{a}{\sqrt3}\), i.e., radius of arc CD

Also, OP = \(\frac{a}{2}\) cot 60º = \(\frac{a}{2\sqrt3}\)

Now are of ΔODC = \(\frac{1}{2}\)x CD x OP = \(\frac{1}{2}\) x a x \(\frac{a}{2\sqrt3}\) = \(\frac{a^2}{4\sqrt3}\)

Area of sector DOC = \(\pi\times\bigg(\frac{a}{\sqrt3}\bigg)^2\times\frac{120}{360}=\frac{\pi{a}^2}{9}\)

[Using Area of secotr = \(\pi{R}^2\big(\frac{\theta}{360°}\big)\)]

∴ Area of segment DLC = Area of sector DOC – Area of ΔDOC

\(\frac{\pi{a}^2}{9}\) - \(\frac{a^2}{4\sqrt3}\)

∴ Total area of the figure = Area of square + Total area of 4 segments 

\(a^2+4\bigg(\frac{\pi{a}^2}{9}-\frac{a^2}{4\sqrt3}\bigg).\)

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