Find the area of the shaded region in the following figure.
It is a square with each side of 4cm and there are two quadrants.

Question

Find the area of the shaded region in the following figure
It is a square with each side of 4cm and there are two
quadrants

Shridhar Kaushik · Accepted Answer

Clearly, OAB is an equilateral triangle and so ∠OAB = ∠OBA = ∠AOB = 60°

∴ ∠CAO = 30°

Now, sector area of OAC is (A₁)

$= \frac{30}{360 °} \times π r^{2} = \frac{1}{12} \times π r^{2} {cm}^{2} ... (1)$

Now sector area of OBA is (A₂)

$= \frac{60}{360 °} \times π r^{2} = \frac{1}{6} \times π r^{2} {cm}^{2} ... (2)$

Also area of equilateral ΔAOB (A₃) $= \frac{\sqrt{3} r^{2}}{4} {cm}^{2}$

∴ Area of minor segment OPA = A₂ – A₃

$= (\frac{π r^{2}}{6} - \frac{\sqrt{3} r^{2}}{4}) {cm}^{2}$

∴ Area of shaded region I is (A) given by

A = A₁ – (area of minor segment OPA)

$= \frac{1}{12} π r^{2} - \frac{π r^{2}}{6} + \frac{\sqrt{3} r^{2}}{4} = \frac{\sqrt{3} r^{2}}{4} - \frac{π r^{2}}{12} = \frac{\sqrt{3}}{4} \times 4 \times 4 - \frac{22}{7 \times 12} \times 4 \times 4 = 4 \sqrt{3} - \frac{22 \times 4}{7 \times 3} = 4 \sqrt{3} - \frac{88}{21} = 6.93 - 4.19 = 2.74 {cm}^{2}$

Also, Area of region II = Area of region I

⇒ Area of required shaded region = 2 (area region of)

⇒ Area of required shaded region = 2 × 2.74 = 5.48 cm²

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Find the area of the shaded region in the following figure. It is a square with each side of 4cm and there are two quadrants.

Find the area of the shaded region in the following figure.
It is a square with each side of 4cm and there are two quadrants.