In the racetrack shown above, regions I and III are semicircular with

Question

https://gmatclub.com/forum/download/file.php?id=47600 In the racetrack shown above, regions I and III are semicircular with radius r. If region II is rectangular and its length is twice its width, what is the perimeter of the track in terms of r ? A. 2r (\pi + 2) B. 2r (\pi + 4) C. 2r (\pi + 8) D. 4r (\pi + 2) E. 4r (\pi + 4) PS35602.01 Quantitative Review 2020 NEW QUESTION 2019-04-26_1746.png

sharathnair14 · Accepted Answer

warrior1991 GMATINSECT GMATestaker
We're given a rectangle and 2 semicircles.
We know for a fact that the diameter of the semicircles is 2r.
And the breadth of the rectangle is equal to the diameter = 2r.

The question says find the perimeter of the track, in Layman terms means the outer boundary length. If you can spare a moment here while you thought of the question, you'll know that while calculating the total peimeter, the breadths of the rectangles aren't to be counted(2r+2r won't count)

Coming to the solution in terms of r,
The perimeter of the 2 semicircles= perimeter of 1 full circle= 2πr
and the perimeter of the rectangle excluding the breadths = 2(4r+2r)-4r= 8r
Adding the two, we get => 2r(π+4)
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Archit3110 · Answer

circumference ; pi*r+ pI*r and for rectangle 
4r+4r
sum all we get
IMO B  2r (\pi + 4)

GMATestaker · Answer

Hi,

A little puzzled here ! perimeter of rectangle is 2(l+w) 
l=2r
w=4r

and always turns out to 2pir +2*6r which is not an answer .Where is the mistake here?

Thanks !

warrior1991 · Answer

chetan2u   Bunuel

The answer should be 2r(π+6). Please help. Am i going wrong somewhere.

The perimeter of rectangle is 2(l+w)
Given l=2w and 
also w=2r since width is the diameter of the circle.

Therefore perimeter of rectangle is 2(2w+w)=6w=12r.

Perimeter of both semicircle is πr+πr=2πr

Answer should be then 2πr+12r =2r(π+6).

warrior1991 · Answer

AjiteshArun  Please validate as to where am i going wrong here.

GMATINSECT · Answer

Struggling with this, any experts to help ?

AjiteshArun · Answer

You've taken the perimeter of the  rectangle  here: The question asks for the perimeter of the  racetrack . This will be 2 sides (top and bottom in the figure) of the rectangle + the circumference of a circle with radius r. That is, we should not consider the dashed lines.

Karmesh · Answer

What are you struggling with? Please elaborate a bit, it makes it much easier to personalise the solution according to your problem so that you may understand better

ScottTargetTestPrep · Answer

We see that the width of the rectangle = diameter of the semicircle = 2r, and the length of the rectangle = 4r.

The perimeter of the track consists of the circumference of one circle and twice the length of the rectangle in the middle. Therefore, the perimeter of the track is:

2πr + 2 x 4r2πr + 8r = 2r(π + 4)

Answer: B

entr0py · Answer

got tripped on this question because I was confused about length vs. width. Normally I would think of width being the horizontal distance, so that's why I got it confused. Upon review I saw that length is twice width, which would mean that it's the longer of the two. Am I supposed to infer from the diagram that length is the longer one then? And for future geometry questions, how should I continue to differentiate between length and width?

Nipunh1991 · Answer

To calculate the length of the track, we need to sum the following:

Length of circumference of Left semi circle + Lengths of the rectangle (top and bottom, no widths to be included) + Length of the circumference of semi circle on the right

The length of each of the following in terms of 'r' are as follows:

(Given : the Length of the rectangle is twice the width i.e 2w=l)
Also width of the rectangle happens to be the diameter of the semicircle = 2r
Therefore 
the width 'w' = 2r
Each length of rectangle = l = 2w = 2( 2r) = 4r
Hence Top and bottom Length combined = 8r

Length of circumference of Left semi circle = Length of circumference of Right semi circle = Pi * r

Summing all the three

Length of circumference of Left semi circle + Lengths of the rectangle (top and bottom, no widths to be included) + Length of the circumference of semi circle on the right 
= pi*r + pi*r + 8r
= 2*pi*r + 8r 
Taking 2r as a common multiple
= 2r (pi + 4)

Hope this helps!

MHIKER · Answer

The perimeter of the two semi-cirlces in the track is  \frac{1}{2} * 2\pi r + \frac{1}{2} * 2\pi r   = 2\pi r

The width of the rectangle  2r   and the length of the rectangle is  4r ; So, the two lengths of the rectangle are  8r

Thus, the perimeter of the track is  2\pi r + 8r   =   2r ( 2\pi  +  4 )

The answer is B

Hoozan · Answer

Similar Question: https://gmatclub.com/forum/the-figure-a ... 13113.html

as19 · Answer

EGMAT::::

Given

In this question, we are given

The diagram of a racetrack consists of two semi-circular regions, with radius r, and one rectangular region.
The length of the rectangular region is twice its width.
To Find

We need to determine

The perimeter of the track, expressed in terms of r.
Approach & Working

The perimeter of each semi-circular region = πr

Therefore, the perimeter of both semi-circular regions together = 2πr
The rectangular region has length twice its width.

Width of the rectangular region = diameter of the semi-circular region = 2r
Hence, length of the rectangular region = 2 * 2r = 4r
Therefore, the perimeter of the whole region = 2πr + 4r + 4r = 2πr + 8r = 2r (π + 4)

Hence, the correct answer is option B.

Answer Choices

bumpbot · Answer

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