An equilateral triangle has side length $8$ . The area of the region containing all points outside the triangle but not more than $3$ units from a point on the triangle is

9 (8 + π)

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8 (9 + π)

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9 (8 + π / 2)

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8 (9 + π / 2)

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Solution

The correct option is A $9 (8 + π)$
The above figure represents geometrical interpretation of the question. The region described contains $3$ rectangles of dimensions $3 X 8,$ and three sectors of circle having radius(r) $3 u n i t s .$
Angle of sector $θ = 360^{\circ} - (90 + 90 + 60)$
$= 120 = \frac{2 Π}{3}$
$A r e a o f s e c t o r = \frac{1}{2} θ . r^{2}$
$R e q u i r e d a r e a = 3 \times$ Area of one rectangle+ $3 \times$ Area of one sector.
$= 3 \times (3 \times 8) + 3. \frac{1}{2} . \frac{2 Π}{3} . 3^{2}$
$∴ R e q u i r e d a r e a = 9 (Π + 8) {u n i t}^{2} .$

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