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# Rectangle ABDC is perfectly inscribed inside a circle and has an area

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Manager
Joined: 19 Dec 2016
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Rectangle ABDC is perfectly inscribed inside a circle and has an area [#permalink]

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02 Aug 2017, 03:55
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Rectangle ABDC is perfectly inscribed inside a circle and has an area of 40 square inches. What is the length of minor arc CD?.

(1) The sum of the measures of angle m and angle n is 60 degrees.
(2) The sum of the measures of angle j and angle n is 90 degrees.
[Reveal] Spoiler: OA

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Senior Manager
Joined: 02 Jul 2017
Posts: 287
GMAT 1: 730 Q50 V38
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area [#permalink]

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12 Sep 2017, 22:04
Refer :
Circle initial : Initial diagram with extra variables for explanation.
Circle Solution : Diagram with all angels and sides marked

Note: Here m writing full explanation so solution might look lengthy but when we solve without writing all the steps it can be done quickly.

Given: ABCD is a rectangle, inscribed in a circle, area 40
=> By figure : AB||CD and AC|| BD
k+n=90 and m+j=90.
Let length and breadth be l and b . so area =40 = l*b
As cord AD and BC make 90 degree at circle. So AD and CB are diagonals. At O they bisect each other. So OA=OB=OC=OD
length of cord AD and BC = $$\sqrt{l^2 +b^2}$$
OC=OD = $$\frac{\sqrt{l^2 +b^2}}{2}$$

Find: Length of Arc CD
=> Arc length = $$2*pi*r* (\frac{angle at center}{360})$$
=> Arc length = $$2*pi*OD* (\frac{angle COD}{360})$$
=> We just need to find length OD and Angle COD

Option 2: The sum of the measures of angle j and angle n is 90 degrees.
so j+n=90
But As given ABCD is a rectangle. Angle D =90. And in triangle CBD sum of all angles =180
=> j+n+90=180 => j+n =90. We know this equation just by question stem. So this option doesn't give us any additional detail.
So Not sufficient.
And this information is already given in question combining 1 and 2 option will not give us any additional detail.
So our Answer is either A or E.

Option 1: The sum of the measures of angle m and angle n is 60 degrees.
m+n=60.
As AB || CD => angle m=n => m=n=30
=> k=j=60
=> by figure: as OC =OD=> n=x=30 and OD=OB => j=y=60
=> angle COD = 120 and Angle DOB=60.
Now as Triangle DOB becomes equilateral triangle. All sides must be equal. => OB=OD=BD
=> $$\frac{\sqrt{l^2 +b^2}}{2} = b$$ => $$l^2 +b^2 =4b^2$$ => $$l^2=3b^2$$ => $$l=\sqrt{3}b$$

as l*b=40 => $$\sqrt{3}b * b = 40$$ => $$b^2= \frac{40}{\sqrt{3}}$$ => $$b = \sqrt{\frac{40}{\sqrt[]3}}$$

Also as radius of circle = r = OD = $$\frac{\sqrt{l^2 +b^2}}{2}$$ = $$\frac{\sqrt{3b^2 +b^2}}{2}$$
$$r = \frac{2b}{2} = b = \sqrt{\frac{40}{\sqrt[]3}}$$

So we got value of angle COD = 120 and value of radius r = $$\sqrt{\frac{40}{\sqrt[]3}}$$

So we can calculate length of Arc CD =$$2*pi*r* (\frac{angle COD}{360})$$

Sufficient

Answer: A
Attachments

File comment: Circle with marked length and angle values. And brief of solution

Circle Solution.jpg [ 2.12 MiB | Viewed 862 times ]

File comment: Initial Circle diagram with additional variables for proper understanding

circle initial.PNG [ 12.91 KiB | Viewed 849 times ]

Intern
Joined: 19 Oct 2015
Posts: 32
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area [#permalink]

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12 Sep 2017, 22:33
@nikkb Question is about finding length of CD. It seems that both A and B are INSUFFICENT.
For me answer is E

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Senior Manager
Joined: 02 Jul 2017
Posts: 287
GMAT 1: 730 Q50 V38
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area [#permalink]

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12 Sep 2017, 23:30
Mujeeb wrote:
Nikkb Question is about finding length of CD. It seems that both A and B are INSUFFICENT.
For me answer is E

Sent from my A0001 using GMAT Club Forum mobile app

Yes i know we have to find length of Arc CD.
Please see the comments i have written on top on the solution. Copy pasting it here again:

Find: Length of Arc CD
=> Arc length = $$2*pi*r* (\frac{angle at center}{360})$$
=> Arc length = $$2*pi*OD* (\frac{angle COD}{360})$$
=> We just need to find length OD and Angle COD

As per formula to calculate length of Arc we should know radius of circle and the angle subtended by the arc at the center.
So here if we calculate radius and Angle COD suptended by arc CD at center O, using formula we can calculate the length of arc.

So as per our calculation Angle COD = 120 and radius = $$\sqrt{\frac{40}{\sqrt[]3}}$$

So we can put this in above formula to calculate length of arc.

Arc CD length = $$2*pi*OD* (\frac{angle COD}{360})$$ = $$2*pi* \sqrt{\frac{40}{\sqrt[]3}} *(\frac{120}{360})$$

Now we don't have to calculate exact value here.

So Answer is A as option 1 alone is sufficient.

Hope m clear.
Intern
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Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area [#permalink]

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12 Sep 2017, 23:33
In this question we should know that the diagonal is diameter hence the angle BDC is 90.

S1- when m+n =60 we get n =30 as both the angle need to be same. So by using tan 30 we can get both BD and CD.

Jyotipravat
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Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area [#permalink]

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19 Oct 2017, 04:41
Another quick way to solve the question:

Lets r be the radius of the cercle, and O be the center.
To get the length of "minor arc CD" we need the measure of "Angle COD" and r.

From stmt(1), m+n=60. n and m has the same measure, hence n=30
we now know that CBD is a (30:60:90) right triangle, which hypthenus measure 2*r .
Therefore, CBD has dimensions (r : sqrt(3)*r : 2*r) --- BD=r and CD=sqrt(3)*r
Area of rectangle ABCD is CD* DB= r * sqrt(3)*r = 40. Hence, we find r.
We also know that triangle COD is isocèles in O and OCD is 30. Hence, "Angle COD" is 180.

Stmt(1) is suff.
Stmt(2) is insuff since it repeats information from the question stem. Nothing new.

Hope it helps...
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Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area   [#permalink] 19 Oct 2017, 04:41
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# Rectangle ABDC is perfectly inscribed inside a circle and has an area

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