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The parallelogram ABCD is inscribed in the circle with centre O as sho

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The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post 28 Mar 2020, 18:42
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GMATBusters’ Quant Quiz Question -7



The parallelogram ABCD is inscribed in the circle with centre O as shown. What is the area of shaded region?
1) Angle OAB = 30 deg and BD = 8
2) ABCD is a rectangle with area 64 root3

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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post 28 Mar 2020, 18:58
The parallelogram ABCD is inscribed in the circle with centre O as shown. What is the area of shaded region?
1) Angle OAB = 30 deg and BD = 8
2) ABCD is a rectangle with area 64 root3

Since angle OAB = 30 deg => angle OBA = 30 deg
Since CB is diameter => CBD is a right angled triangle
Angle OBD = angle ODB = 60 deg which mean triangle ODB is an equilateral triangle
With length of BD = 8 => we know the area of equilateral triangle
We know area of triangle AOC
Since we know the radius = OB = BD = 8 => We would know the area of the circle

Area of the shaded region would thus be known = (area of circle/6) - Area of equilateral triangle
Sufficient

ii) ABCD is a rectangle with area => Insufficient as we don't know the lengths of individual sides

Answer - A
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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post 28 Mar 2020, 21:41
Given: The parallelogram ABCD is inscribed in the circle with centre O as shown.

Asked: What is the area of shaded region?

Since diagonals of parallelogram ABCD intersect at centre of the circle at O , AD & BC are diameters of the circle and angle CAB = angle CDB = angle ABD = angle ACD =90
ABCD is a rectangle.
If radius of the circle r and angles AOC = x are known ;
Area of the shaded region = \(\pi r^2 * (x/360) - 1/2 r^2 sin x\)

1) Angle OAB = 30 deg and BD = 8
Since angle OAB = 30 ; angle AOB = 180 - 2*30 = 120; angle AOC = 180 - angle AOB = 180 - 120 = 60 ; Triangle AOC is equilateral triangle ; BD = AC = 8 = r = AO = )C
Since r = 8 and angle AOC = 60
Area of the shaded region = \(\pi 8^2 * (\frac{60}{360}) - \frac{1}{2}* 8^2 sin 60 = 64 (\frac{\pi}{6} - \frac{\sqrt{3}}{4})\)
SUFFICIENT

2) ABCD is a rectangle with area 64 root3
\(AC * AB = 64 \sqrt{3}\)
Let x & y be sides of the rectangle
\(x*y = 64\sqrt{3}\)
\(x^2 + y^2 = 4r^2\)
It is NOT possible to derive r and angle AOC from information provided since various combinations are possible
NOT SUFFICIENT

IMO A
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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post 28 Mar 2020, 22:16
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The parallelogram ABCD is inscribed in the circle with centre O as shown. What is the area of shaded region?

1) Angle OAB = 30 deg and BD = 8,
In triangle OAB, <OAB= 30, <OBA = 30 as OA and OB are radius and have equal length.
So, < AOB = 120. Therefore, <COD also 120 and <AOC = 60 and <BOD = 60.
BD =8 and AC = 8 as opposite lengths are equal in parallelogram.
Triangle AOC forms equilateral triangle as OA=OC and <AOC = 60. So, all sides have 8 length (= radius of circle).
Now, we can find the area of shaded region,
<AOC/360 = Area of sector AOC/Circumferance of circle. (circumferance of circle = 2*3.14* 8 = 16*3.14)
30/360 = Area of sector AOC/16*3.14
Area of sector = unique no.
Now we can deduct the area of triangle AOC from the Area of sector find the area of shaded region.
Sufficient.

2) ABCD is a rectangle with area 64 root3
Area of rectangle ABCD = 64 root3
AC * CD = 64 root3
AC^2 * CD^2 = 64*64*3
Area of triangle ACD (right angle triangle) = half of area of rectangle = 32 root3
1/2 * AC * CD = 32 root3
< ACD = 90 as ABCD is rectangle but we can't find the angle of AOC. So, no unique value of AO and OC nor respective angles. So, insufficient.

Ans. A.
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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post 28 Mar 2020, 23:04
The parallelogram ABCD is inscribed in the circle with centre O as shown. What is the area of shaded region?
1) Angle OAB = 30 deg and BD = 8
2) ABCD is a rectangle with area 64 root3


the only parallelogram that can be inscribed in a circle is a rectangle. To find the area, we would need the angle of the sector, and the radius of the circle/chord length

Statement (1) tells us OAB is 30, from which we can define the inscribed sector angle is 60deg. combined with eht BD = 8, this is sufficient geometric definition to learn everything we need about the overall circle area (radius 8), sector area (1/6 total circle area), and segment area (subtract triangular area from sector area!) --> sufficient --> eliminate (b), (c), (e)

Statement (2) is immediately insufficient. If there is only one inscribed rectangle with that total area possible, we don't know which segment (relatively speaking, on the "long side" or "short side") of the rectangle we're needing to calculate --> insufficient --> eliminate (d) --> answer is (a)
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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post Updated on: 29 Mar 2020, 20:53
Given: The parallelogram ABCD is inscribed in the circle with centre O as shown.

Asked: What is the area of shaded region?

Since diagonals of parallelogram ABCD intersect at centre of the circle at O , AD & BC are diameters of the circle and angle CAB = angle CDB = angle ABD = angle ACD =90
ABCD is a rectangle.
If radius of the circle r and angles AOC = x are known ;
Area of the shaded region = \(\pi r^2 * (x/360) - 1/2 r^2 sin x\)

1) Angle OAB = 30 deg and BD = 8
Since angle OAB = 30 ; angle AOB = 180 - 2*30 = 120; angle AOC = 180 - angle AOB = 180 - 120 = 60 ; Triangle AOC is equilateral triangle ; BD = AC = 8 = r = AO = )C
Since r = 8 and angle AOC = 60
Area of the shaded region = \(\pi 8^2 * (\frac{60}{360}) - \frac{1}{2}* 8^2 sin 60 = 64 (\frac{\pi}{6} - \frac{\sqrt{3}}{4})\)
SUFFICIENT

2) ABCD is a rectangle with area 64 root3
\(AC * AB = 64 \sqrt{3}\)
Let x & y be sides of the rectangle
\(x*y = 64\sqrt{3}\)
Since there are many possibilities for x & y
NOT SUFFICIENT

IMO A
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Originally posted by Kinshook on 29 Mar 2020, 01:46.
Last edited by Kinshook on 29 Mar 2020, 20:53, edited 3 times in total.
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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post 29 Mar 2020, 04:56
we have 2 angles and a side . we can get all the values for the diagram . so only a option is enough to calculate
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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post 29 Mar 2020, 06:20
(A) Statemenr 1 alone is sufficient
BD=8 units
Statement-1
OAB=30=>DOB=60 (Angle subtended by a chord at the center is double that which it subtends at the circle boundary)
Since OD=OB=radius, this implies DOB is an equilateral triangle with all Angles 60.
That implies that all sides are equal (including radius)=8
With radius and angle known, we can find area of shaded = area of sector - area of BOD
With statement-1 we can thus find area
However with statement we cannot quantify length or breadth and hence that isn't sufficient

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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho  [#permalink]

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New post 29 Mar 2020, 07:24
Not sure about the solution.

(1) + (2)
\( \triangle\)AOC = \(\frac{1}{2}*8*4\sqrt{3} \)
= \(16\sqrt{3} \)

Now since it is a rectangle, by Pythagoras theorem we can find out the diameter = 16

Area of circle = \(\pi * 8^2\) = 200.1
Area subtended by 60 degrees = \(\frac{360}{60} * 200.1\) = 1200.6

Area AC = 1200.6 - \(16\sqrt{3}\)

Answer: C
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Re: The parallelogram ABCD is inscribed in the circle with centre O as sho   [#permalink] 29 Mar 2020, 07:24

The parallelogram ABCD is inscribed in the circle with centre O as sho

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