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What is the area of the shaded region above, if ABCD

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What is the area of the shaded region above, if ABCD  [#permalink]

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New post 11 Jan 2014, 07:16
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A
B
C
D
E

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  55% (hard)

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65% (02:00) correct 35% (01:56) wrong based on 262 sessions

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Attachment:
File comment: Geometry
ABCD.png
ABCD.png [ 20.95 KiB | Viewed 8485 times ]

What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?

(1) The radius of the circle with center O is 4.
(2) The area of triangle ADC is 32.

OE
To find area of shaded region, you need to find area of triangle ACD, and subtract area of unshaded region inside this triangle from it. To find area of unshaded region, you need to first draw in a line perpendicular to diameter AB of circle from center O to diagonal AC. Call this point E. This point is also one of endpoints of arc AFE.
Attachment:
File comment: ABCD2
ABCD2.png
ABCD2.png [ 19.07 KiB | Viewed 7238 times ]

Arc AFE and radii OA and OE bound a sector of circle O: this sector is one quarter of entire area of circle. Now, if you take area of small triangle AEO, and subtract this area from area of sector OAFE, then you have area of unshaded portion of triangle ADC. So, to solve this question, you need to find area of triangle ADC, one quarter of area of circle O, and area of small triangle AEO. To find these three areas, you need radius of circle, or length of one side of square.

(1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient.

(2): able to find length of height and width (which are equal to each other) of triangle ADC.
Let's say that both base and height of triangle ADC = x.
32 = (½)x^2 → x = 8
You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius.
Sufficient
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Re: What is the area of the shaded region above, if ABCD  [#permalink]

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New post 11 Jan 2014, 07:26
1
goodyear2013 wrote:
Attachment:
ABCD.png

What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?

(1) The radius of the circle with center O is 4.
(2) The area of triangle ADC is 32.

OE
To find area of shaded region, you need to find area of triangle ACD, and subtract area of unshaded region inside this triangle from it. To find area of unshaded region, you need to first draw in a line perpendicular to diameter AB of circle from center O to diagonal AC. Call this point E. This point is also one of endpoints of arc AFE.
Attachment:
ABCD2.png

Arc AFE and radii OA and OE bound a sector of circle O: this sector is one quarter of entire area of circle. Now, if you take area of small triangle AEO, and subtract this area from area of sector OAFE, then you have area of unshaded portion of triangle ADC. So, to solve this question, you need to find area of triangle ADC, one quarter of area of circle O, and area of small triangle AEO. To find these three areas, you need radius of circle, or length of one side of square.

(1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient.

(2): able to find length of height and width (which are equal to each other) of triangle ADC.
Let's say that both base and height of triangle ADC = x.
32 = (½)x^2 → x = 8
You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius.
Sufficient


Image
What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?

Actually there is no need for any calculations. From both statements we have everything "fixed", only one case possible where we know everything possible.

(1) The radius of the circle with center O is 4 --> diameter = side = 8. We have "fixed" circle and square. We can find anything there. Sufficient.

(2) The area of triangle ADC is 32 --> 1/2*AD^2=32 --> AD=8 --> AD = diameter = 8. The same info as above. Sufficient.

Answer: D.
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Re: What is the area of the shaded region above, if ABCD  [#permalink]

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New post 11 Jan 2014, 15:55
Bunuel wrote:
goodyear2013 wrote:
Attachment:
ABCD.png

What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?

(1) The radius of the circle with center O is 4.
(2) The area of triangle ADC is 32.

OE
To find area of shaded region, you need to find area of triangle ACD, and subtract area of unshaded region inside this triangle from it. To find area of unshaded region, you need to first draw in a line perpendicular to diameter AB of circle from center O to diagonal AC. Call this point E. This point is also one of endpoints of arc AFE.
Attachment:
ABCD2.png

Arc AFE and radii OA and OE bound a sector of circle O: this sector is one quarter of entire area of circle. Now, if you take area of small triangle AEO, and subtract this area from area of sector OAFE, then you have area of unshaded portion of triangle ADC. So, to solve this question, you need to find area of triangle ADC, one quarter of area of circle O, and area of small triangle AEO. To find these three areas, you need radius of circle, or length of one side of square.

(1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient.

(2): able to find length of height and width (which are equal to each other) of triangle ADC.
Let's say that both base and height of triangle ADC = x.
32 = (½)x^2 → x = 8
You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius.
Sufficient


Image
What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?

Actually there is no need for any calculations. From both statements we have everything "fixed", only one case possible where we know everything possible.

(1) The radius of the circle with center O is 4 --> diameter = side = 8. We have "fixed" circle and square. We can find anything there. Sufficient.

(2) The area of triangle ADC is 32 --> 1/2*AD^2=32 --> AD=8 --> AD = diameter = 8. The same info as above. Sufficient.

Answer: D.


Bunuel, Can you please explain why if you only have the radius you can find anything. How would you get the AF part?

I knew that Statement 1 and 2 gave the same information, but couldn't figure out how to find the white portion in TriangleADC

Thanks
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Re: What is the area of the shaded region above, if ABCD  [#permalink]

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New post 12 Jan 2014, 05:36
b00gigi wrote:
Bunuel wrote:
goodyear2013 wrote:
Attachment:
ABCD.png

What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?

(1) The radius of the circle with center O is 4.
(2) The area of triangle ADC is 32.

OE
To find area of shaded region, you need to find area of triangle ACD, and subtract area of unshaded region inside this triangle from it. To find area of unshaded region, you need to first draw in a line perpendicular to diameter AB of circle from center O to diagonal AC. Call this point E. This point is also one of endpoints of arc AFE.
Attachment:
ABCD2.png

Arc AFE and radii OA and OE bound a sector of circle O: this sector is one quarter of entire area of circle. Now, if you take area of small triangle AEO, and subtract this area from area of sector OAFE, then you have area of unshaded portion of triangle ADC. So, to solve this question, you need to find area of triangle ADC, one quarter of area of circle O, and area of small triangle AEO. To find these three areas, you need radius of circle, or length of one side of square.

(1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient.

(2): able to find length of height and width (which are equal to each other) of triangle ADC.
Let's say that both base and height of triangle ADC = x.
32 = (½)x^2 → x = 8
You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius.
Sufficient


Image
What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?

Actually there is no need for any calculations. From both statements we have everything "fixed", only one case possible where we know everything possible.

(1) The radius of the circle with center O is 4 --> diameter = side = 8. We have "fixed" circle and square. We can find anything there. Sufficient.

(2) The area of triangle ADC is 32 --> 1/2*AD^2=32 --> AD=8 --> AD = diameter = 8. The same info as above. Sufficient.

Answer: D.


Bunuel, Can you please explain why if you only have the radius you can find anything. How would you get the AF part?

I knew that Statement 1 and 2 gave the same information, but couldn't figure out how to find the white portion in TriangleADC

Thanks


That's the beauty of DS: we don't need to solve if we know that we CAN solve. We can solve because we know the radius of the circle (so we know everything about it) and we know the side of the square (so we know everything about it). If still interested in solving you can check under the spoiler in the initial post.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: What is the area of the shaded region above, if ABCD  [#permalink]

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New post 27 Jan 2015, 11:42
D--

One subtle hint....both the options A and B pointing towards same conclusion.....so can be either D or E....

Like the solution provided in spoiler....either is enough
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Re: What is the area of the shaded region above, if ABCD  [#permalink]

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Re: What is the area of the shaded region above, if ABCD &nbs [#permalink] 10 Sep 2018, 19:13
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