A circle having radius 4cm contains a chord of length 4cm and subtends an angle of 60 degrees. Find the area (in $c m^{2}$ )of the minor segment of the chord.

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1.5

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None of these

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Solution

The correct option is B
1.5

Area of sector POQ= $\frac{60}{360}$ × $π 4^{2}$ = 8.4 $c m^{2}$

In triangle OSQ which is right angled at S, $O Q^{2}$ = $S Q^{2}$ + $O S^{2}$

16 = 4 + $O S^{2}$ OS = 2 $\sqrt{3}$

Area of triangle POQ = $\frac{1}{2}$ × base × height=

$\frac{1}{2}$ × 4 × 2 $\sqrt{3}$

= 6.9 $c m^{2}$ Area of segment PSQR = 8.4 – 6.9 = 1.5 $c m^{2}$

Area of triangle POQ = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × 4 × 2 $\sqrt{3}$ = 6.9 $c m^{2}$

Area of segment PSQR = 8.4 – 6.9 = 1.5 $c m^{2}$

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A circle having radius 4cm contains a chord of length 4cm and subtends an angle of 60 degrees. Find the area (in cm2)of the minor segment of the chord.

A circle having radius 4cm contains a chord of length 4cm and subtends an angle of 60 degrees. Find the area (in $c m^{2}$ )of the minor segment of the chord.