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23 Nov 2017, 01:22
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B
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Difficulty:
75%
(hard)
Question Stats:
58% (03:17) correct
42%
(03:29)
wrong
based on 66
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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 [ 13.95 KiB | Viewed 31871 times ]
23 Nov 2017, 17:41
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Bunuel
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.
\(3π + 12 - 8 + 6 = 10 + 3π\)
23 Nov 2017, 23:41
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Bunuel
Perimeter of shaded region = Perimeter of quarter circle SRT - (Length + Width of rectangle) + Diagonal of rectangle (AC)
The diagonal of the rectangle is same as the radius of the circle i.e. 6.
Perimeter of quarter circle SRT = 6 + 6 + (1/4)*2π*6 = 12 + 3π
Perimeter of shaded region = 12 + 3π - 8 + 6 = 10 + 3π
Answer (B)
