Bunuel wrote:

In the figure above, arc SBT is one quarter of a circle with center R and radius 6. If the length plus the width of rectangle ABCR is 8, then the perimeter of the shaded region is

(A) 8 + 3π

(B) 10 + 3π

(C) 14 + 3π

(D) 1 + 6π

(E) 12 + 6π

Attachment:

2017-11-23_1215_001.png

It should be

B.

Length of Arc is quarter of circumference of circle so

\(Arc = 2πr/4 = 2π6/4 = 3π\)

Rectangle ABCR is sum of two triangles \(\triangle ABC\) and \(\triangle ARC\) where diagonals \(RB\) and \(AC\) are equal to radius of the Circle. If it were a regular quarter circle, perimeter would have been 3π + 2*radius. However, It appears the shaded region just subtracts length and breadth of the rectangle (given as 8) from the two radius and bypasses it with the diagonal of the rectangle instead.

\(3π + 12 - 8 + 6 = 10 + 3π\)

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