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# In the figure above, arc SBT is one quarter of a circle with center R

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Joined: 02 Sep 2009
Posts: 59722
In the figure above, arc SBT is one quarter of a circle with center R  [#permalink]

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23 Nov 2017, 01:22
00:00

Difficulty:

85% (hard)

Question Stats:

47% (02:54) correct 53% (03:10) wrong based on 32 sessions

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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 8199 times ]

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Joined: 14 Oct 2015
Posts: 243
GPA: 3.57
Re: In the figure above, arc SBT is one quarter of a circle with center R  [#permalink]

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23 Nov 2017, 17:41
1
1
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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Re: In the figure above, arc SBT is one quarter of a circle with center R  [#permalink]

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23 Nov 2017, 23:41
3
1
1
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

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

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Re: In the figure above, arc SBT is one quarter of a circle with center R  [#permalink]

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22 Nov 2019, 20:47
1
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

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

Hi, Could you please explain how diagonal of the rectangle = radius of the circle?
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Re: In the figure above, arc SBT is one quarter of a circle with center R  [#permalink]

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22 Nov 2019, 21:17
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

Perimeter of the shaded region = Perimeter of quarter circle + radius of the circle - length and width of the rectangle = 6 + 6 + 2π6*1/4 + 6 -8 = 18-8 + 3π = 10 + 3π

IMO B
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Re: In the figure above, arc SBT is one quarter of a circle with center R  [#permalink]

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23 Nov 2019, 08:53
1
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9876
Location: Pune, India
Re: In the figure above, arc SBT is one quarter of a circle with center R  [#permalink]

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24 Nov 2019, 22:12
dips1122 wrote:
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

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

Hi, Could you please explain how diagonal of the rectangle = radius of the circle?

We know that in a rectangle, both diagonals are equal in length.
Look at the other diagonal RB of the rectangle. It is obviously the radius of the quarter circle (from the centre of the circle R to a point on the circle B). Hence the diagonal of the rectangle is the same length as the radius of the circle i.e. 6.
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Karishma
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Joined: 16 Oct 2010
Posts: 9876
Location: Pune, India
Re: In the figure above, arc SBT is one quarter of a circle with center R  [#permalink]

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24 Nov 2019, 22:19
shahidomer77 wrote:

Perimeter of shaded region = SACTBS = AS + SB + BT + TC + AC

To get ASBTC (AS + SB + BT + TC), we find the complete perimeter of the quarter circle and subtract AR and RC out of it.

ASBTC = 12 + 3π - (AR - RC) = 12 + 3π - 8

But we need to add the length of AC still. As discussed in my comment above, AC = 6

So SACTBS = ASBTC + AC = 12 + 3π - 8 + 6
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Karishma
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Re: In the figure above, arc SBT is one quarter of a circle with center R   [#permalink] 24 Nov 2019, 22:19
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