The area of the triangle formed by the points given in the polar coordinate system $(1, \frac{π}{6})$ , $(2, \frac{π}{3})$ and $(3, \frac{π}{2})$ , is

\frac{3 \sqrt{3} - 8}{4}

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\frac{8 - 3 \sqrt{3}}{4}

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\frac{8 + 3 \sqrt{3}}{4}

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\frac{8 + 3 \sqrt{2}}{4}

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Solution

The correct option is B $\frac{8 - 3 \sqrt{3}}{4}$
Given points $(1, \frac{π}{6})$ , $(2, \frac{π}{3})$ and $(3, \frac{π}{2})$
Area of triangle, $△ = \frac{1}{2} | r_{1} r_{2} sin (θ_{2} - θ_{1}) + r_{2} r_{3} sin (θ_{3} - θ_{2}) + r_{3} r_{1} sin (θ_{1} - θ_{3}) |$
$= \frac{1}{2} ∣ ∣ ∣ 2 sin (\frac{π}{6}) + 6 sin (\frac{π}{6}) + 3 sin (- \frac{π}{3}) ∣ ∣ ∣$
$= \frac{1}{2} ∣ ∣ ∣ \frac{8 - 3 \sqrt{3}}{2} ∣ ∣ ∣$
$= \frac{8 - 3 \sqrt{3}}{4}$

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