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The circular base of an above-ground swimming pool lies in a

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

(1) The base has area 250 square feet.
(2) The center of the base is 20 feet from point C.

Data Sufficiency
Question: 99
Category: Geometry Circles
Page: 160
Difficulty: 650


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

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The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 18 Feb 2014, 02:51
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The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

Consider the diagram below:
Image
Notice that we are asked to find the length of QA, or the radius of the circular base.

(1) The base has area 250 square feet --> \(area=\pi{r^2}=250\) --> we can find r. Sufficient.

(2) The center of the base is 20 feet from point C. Triangle CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

So, this statement is not sufficient to get QA.

Answer: A.

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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 18 Feb 2014, 04:10
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IMO A,

Q How far from the center of the pool's base is point A? i.e. we have to find the radius of the pool

1) area = 250 sq feet. we can use pi(r^2)=250 to find the radius of the pool hence A is sufficient

2) let the center be o, we have OC=20, by using Pythagoras theorem in right angle triangle OCA, (as CA is tangent to the circular base )

we have (20)^2= (OA)^2+(CA)^2
this equation can have infinitely many solutions hence 2 alone is not sufficient
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 18 Feb 2014, 04:49
1
Basically, we need to find the radius of the given circle.

St1: The base has area 250 square feet. Sufficient.

St2: The center of the base is 20 feet from point C.

Let the center be O. This implies OC = 20. Since a tangent is perpendicular with the radius of a circle, we will have triangle OAC as a right triangle, right angled at A. OC is the hypotenuse. Even using Pythagoras theorem, we cannot get a definite value for the radius. Not sufficient.

Answer (A).
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 22 Feb 2014, 06:03
Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

Consider the diagram below:
Image
Notice that we are asked to find the length of QA, or the radius of the circular base.

(1) The base has area 250 square feet --> \(area=\pi{r^2}=250\) --> we can find r. Sufficient.

(2) The center of the base is 20 feet from point C. Triangle CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

So, this statement is not sufficient to get QA.

Answer: A.
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 18 May 2014, 09:31
We can see that the distance of the center of the pool's base, O and the point A is nothing but the radius of the circular base. Hence, we have to find find the radius of the base.

Statement 1: We know the area of the circular base.
Hence, we can determine the radius of the base.

Sufficient

Statement 2: OC = 20
This doesn't tell us anything about the radius.

Not sufficient

The correct answer is A.
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 27 May 2014, 14:55
Bunuel wrote:
Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

Consider the diagram below:
Image
Notice that we are asked to find the length of QA, or the radius of the circular base.

(1) The base has area 250 square feet --> \(area=\pi{r^2}=250\) --> we can find r. Sufficient.

(2) The center of the base is 20 feet from point C. Triangle CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

So, this statement is not sufficient to get QA.

Answer: A.



Hi Bunuel,

For the sake of argument(I know that this doesn't provide any additional value to help solve this problem) -- how can we assume that BCQ and ACQ are equal? I realize that both lines are tangent to the circle but doesn't that mean that the line is perpendicular to the center of the circle. Theoretically, that has infinite points, doesn't it?

Thanks
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 28 May 2014, 00:24
russ9 wrote:
Bunuel wrote:
Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

Consider the diagram below:
Image
Notice that we are asked to find the length of QA, or the radius of the circular base.

(1) The base has area 250 square feet --> \(area=\pi{r^2}=250\) --> we can find r. Sufficient.

(2) The center of the base is 20 feet from point C. Triangle CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

So, this statement is not sufficient to get QA.

Answer: A.



Hi Bunuel,

For the sake of argument(I know that this doesn't provide any additional value to help solve this problem) -- how can we assume that BCQ and ACQ are equal? I realize that both lines are tangent to the circle but doesn't that mean that the line is perpendicular to the center of the circle. Theoretically, that has infinite points, doesn't it?

Thanks


Two important properties:
1. The tangent line is always at the 90 degree angle (perpendicular) to the radius of a circle.
2. Tangent segments to a circle from the same external point are congruent.


Hope it helps.
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 01 Nov 2014, 13:03
Bunuel wrote:
Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

Consider the diagram below: Image
Notice that we are asked to find the length of QA, or the radius of the circular base.

(1) The base has area 250 square feet --> \(area=\pi{r^2}=250\) --> we can find r. Sufficient.

(2) The center of the base is 20 feet from point C. Triangle CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

So, this statement is not sufficient to get QA.

Answer: A.


Hi Bunuel,

A few follow up questions to statement 2:

1) If we know that the segments BC and AC are equal (one of the properties of a tangent line that intersects a circle), can't we assume that the angle QAC is a 45.45.90 ? I would imagine that since QA = QB and since they are both 90, they will intersect the circle symmetrically and therefore create a 45 45 90. Am I inferring too much here?

2) If we know that hypotenuse is 20, I was under the impression that a 20 hyp and a 90 degree angle ALWAYS results in a 12:16:20?

Thanks!
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 02 Nov 2014, 05:19
russ9 wrote:
Bunuel wrote:
Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

Consider the diagram below: Image
Notice that we are asked to find the length of QA, or the radius of the circular base.

(1) The base has area 250 square feet --> \(area=\pi{r^2}=250\) --> we can find r. Sufficient.

(2) The center of the base is 20 feet from point C. Triangle CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

So, this statement is not sufficient to get QA.

Answer: A.


Hi Bunuel,

A few follow up questions to statement 2:

1) If we know that the segments BC and AC are equal (one of the properties of a tangent line that intersects a circle), can't we assume that the angle QAC is a 45.45.90 ? I would imagine that since QA = QB and since they are both 90, they will intersect the circle symmetrically and therefore create a 45 45 90. Am I inferring too much here?

2) If we know that hypotenuse is 20, I was under the impression that a 20 hyp and a 90 degree angle ALWAYS results in a 12:16:20?

Thanks!


1. Yes, CA = CB. You are right about it. But this does not mean that CQA or CQB are 45-45-90 triangles. Consider this, by moving point C triangles CQA and CQB change, so only one position of C would give 45-45-90 triangles, while other positions of C will give other types of right triangles.

2. This is explained in the solution. Knowing that hypotenuse is 20 does not necessarily mean that we have a 12:16:20 right triangle. For more on this trap check the following questions:
what-is-the-area-of-parallelogram-abcd-111927.html
the-circular-base-of-an-above-ground-swimming-pool-lies-in-a-167645.html
figure-abcd-is-a-rectangle-with-sides-of-length-x-centimete-48899.html
in-right-triangle-abc-bc-is-the-hypotenuse-if-bc-is-13-and-163591.html
m22-73309-20.html
points-a-b-and-c-lie-on-a-circle-of-radius-1-what-is-the-84423.html
if-vertices-of-a-triangle-have-coordinates-2-2-3-2-and-82159-20.html
if-p-is-the-perimeter-of-rectangle-q-what-is-the-value-of-p-135832.html
if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html
what-is-the-area-of-rectangular-region-r-105414.html
what-is-the-perimeter-of-rectangle-r-96381.html
pythagorean-triples-131161.html
given-that-abcd-is-a-rectangle-is-the-area-of-triangle-abe-127051.html
m13-q5-69732-20.html#p1176059
m20-07-triangle-inside-a-circle-71559.html
what-is-the-perimeter-of-rectangle-r-96381.html
what-is-the-area-of-rectangular-region-r-166186.html
if-distinct-points-a-b-c-and-d-form-a-right-triangle-abc-129328.html

Hope this helps.
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 03 Jul 2015, 06:46
Bunuel wrote:
Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

Consider the diagram below:
Attachment:
Untitled.png
Notice that we are asked to find the length of QA, or the radius of the circular base.

(1) The base has area 250 square feet --> \(area=\pi{r^2}=250\) --> we can find r. Sufficient.

(2) The center of the base is 20 feet from point C. Triangle CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

So, this statement is not sufficient to get QA.

Answer: A.


If the question had a different context, and it said the sides had to be integers, would statement 2 be sufficient ?
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 02 May 2016, 07:28
bluepulaski1 wrote:
Bunuel wrote:
Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

Consider the diagram below:
Attachment:
Untitled.png
Notice that we are asked to find the length of QA, or the radius of the circular base.

(1) The base has area 250 square feet --> \(area=\pi{r^2}=250\) --> we can find r. Sufficient.

(2) The center of the base is 20 feet from point C. Triangle CQA IS a right triangle, because the tangent line (CA) is always at the 90 degree angle (perpendicular) to the radius (QA) of a circle.

So, we have that CQ=hypotenuse=20. BUT, knowing that hypotenuse equals to 20 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 12:16:20. Or in other words: if \(x^2+y^2=20^2\) DOES NOT mean that \(x=12\) and \(y=16\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=20^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=12\) and \(y=16\).

For example: \(x=1\) and \(y=\sqrt{399}\) or \(x=2\) and \(y=\sqrt{396}\)...

So, this statement is not sufficient to get QA.

Answer: A.


If the question had a different context, and it said the sides had to be integers, would statement 2 be sufficient ?

Yes by declaring the sides are integers only we will eliminate all other possible combinations and will be left with only integer combination of 12 and 16,which gives 20 as hypotenuse.
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 31 May 2017, 05:56
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

(1) The base has area 250 square feet.
(2) The center of the base is 20 feet from point C.

Data Sufficiency
Question: 99
Category: Geometry Circles
Page: 160
Difficulty: 650


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!


Attachment:
Untitled.png


The goal of the problem is to find the distance between the center of the pool to point A, a point in the edge of the pool. In other words, what is the radius of the pool?

Statement 1) The base has area 250 square feet.

A = pi*R^2

250 = pi*R^2

Sufficient.

Statement 2) Center of the base is 20 feet from point C.

Whenever a point is tangent to a circle, it is perpendicular to the radius. In this case, it creates a right triangle with hypotenuse 20 from C to O, the center of the circle. However, we don't know the dimensions of the at least one of the other sides, so we cannot find the radius. Insufficient.
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Re: The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 11 Jul 2020, 03:30
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Image
The circular base of an above-ground swimming pool lies in a level yard and just touches two straight sides of a fence at points A and B, as shown in the figure above. Point C is on the ground where the two sides of the fence meet. How far from the center of the pool's base is point A?

(1) The base has area 250 square feet.
(2) The center of the base is 20 feet from point C.

Data Sufficiency
Question: 99
Category: Geometry Circles
Page: 160
Difficulty: 650


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!


Attachment:
Untitled.png


Wanted: radius?

1) Area given => Can get radius
Sufficient

2) Let centre be at O, then OC = 20

r^2 + CB^2 = 400
Don't know CB
Not sufficient

ANSWER: A
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The circular base of an above-ground swimming pool lies in a  [#permalink]

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New post 03 Aug 2020, 07:51
I have a question about statement 2. If we build polygon OABC, then sides OA and OB are equal since both of them are radiuses and angles CBO and CAO are 90 degrees , therefore OABC is a square. Then CO is diagonal and equals square root of ( 2*(one side of square^2) = 20 and statement 2 is sufficient. Or does the property two adjacent sides equal - polygon is square doesn't apply to all polygons?
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The circular base of an above-ground swimming pool lies in a   [#permalink] 03 Aug 2020, 07:51

The circular base of an above-ground swimming pool lies in a

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