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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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Hi pacifist85,

While "adding in" a radius can be a useful 'step' in dealing with a multiple-shape Geometry question, it won't actually help us here.

With the information in Fact 2, we know that we have a right triangle with a hypotenuse of 18, but we don't know ANYTHING else.

With that info, we have A^2 + B^2 = 18^2, but we don't know the actual values of A and B (which we need to figure out the area). If we had info about the other angles, then we COULD figure those sides out though (since we'd have a relationship among the 3 sides based on the angles).

Adding in a radius won't help us figure out any of the other angles (in the big triangle or in either of the smaller triangles).

To figure out the area of the triangle, we need a "base" and a "height." We COULD set the 18 as the base, but we have no way of determining the height without additional information (at least one more of the sides or one of the non-90 degree angles). As such, Fact 2 is INSUFFICIENT.

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Senior Manager  Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 412 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) For the triangle shown, where A, B and C are all points on [#permalink] Show Tags Hi Rich, I understand eveything you said. Actually, I wasn't assuming any of this. I added the image to illustrate what I thought. So, after drawing it, I realised the mistake in my line of reasoning. The problem is that I cannot be sure which one the hypotenuse is, in any of the 2 smaller triangles, so using the pythagorean I can find the third side. Right? Because all I know is that AC + CB >18 and that 18> CB - AC. But even with this information, I cannot figure out which one of the sides has the bigger or smaller length. I was focusing on the lower triangle, because it seems that CB is greater than the radius, but I cannot be sure. So, "it seems" is not enough. Attachments Untitled.png [ 6.63 KiB | Viewed 1641 times ] EMPOWERgmat Instructor V Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14338 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: For the triangle shown, where A, B and C are all points on [#permalink] Show Tags Hi pacifist85, A hypoteneuse ONLY exists in right triangles, so neither of the two small triangles actually has a hypoteneuse. When dealing with Geometry questions on the GMAT, it's often really helpful to draw the physical shapes (as opposed to just staring at words or sticking to just the math formulas involved). You'll be far more likely to make 'connections' and the proper deductions when you can "see" the shapes and your work, so I encourage you to continue doing what you did here - draw the pictures and get the points. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Rich Cohen Co-Founder & GMAT Assassin Follow Special Offer: Save$75 + GMAT Club Tests Free
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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EMPOWERgmatRichC wrote:
Hi pacifist85,

A hypoteneuse ONLY exists in right triangles, so neither of the two small triangles actually has a hypoteneuse.

When dealing with Geometry questions on the GMAT, it's often really helpful to draw the physical shapes (as opposed to just staring at words or sticking to just the math formulas involved). You'll be far more likely to make 'connections' and the proper deductions when you can "see" the shapes and your work, so I encourage you to continue doing what you did here - draw the pictures and get the points.

GMAT assassins aren't born, they're made,
Rich

Hmmmm... a very good point that did not even occur to me... OK I will pretend that I had a mini stroke an move on!
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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enigma123 wrote:
Attachment:
Untitled.png
For the triangle shown, where A, B and C are all points on a circle, and line segment AB has length 18, what is the area of triangle ABC?

(1) Angle ABC measures 30°.
(2)The circumference of the circle is $$18\pi$$.

Choice (1) by itself is insufficient. However, there is an equation to find the area of a triangle

Area of a triangle, A = P * r /2
P = perimeter of the inscribed circle
r = radius of the inscribed circle

Won't B alone suffice to reach the answer?
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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enigma123 wrote:
Attachment:
Untitled.png
For the triangle shown, where A, B and C are all points on a circle, and line segment AB has length 18, what is the area of triangle ABC?

(1) Angle ABC measures 30°.
(2)The circumference of the circle is $$18\pi$$.

To be honest it's a very easy question, if you don't assume your decisions based just on the drawing ! -- > It's a DS question, so don't assume anything.
(1) We have ONLY one side and one angle, it's not sufficient to calculate the area of a triamgle
(2) C=18pi, this tells us that the diameter is equal to 18 and that it's a right triangle, BUT it's still not sufficient to calculate an area with ONLY one side and one angle
(1) + (2) It's a 30-60-90 Triangle, thus we can calculate an are with the given information (all 3 angles + Hypotenuse =18)
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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In order to find the area of the triangle, we need to find the lengths of a base and its associated height. Our strategy will be to prove that ABC is a right triangle, so that CB will be the base and AC will be its associated height.

(1) INSUFFICIENT: We now know one of the angles of triangle ABC, but this does not provide sufficient information to solve for the missing side lengths.

(2) INSUFFICIENT: Statement (2) says that the circumference of the circle is 18 . Since the circumference of a circle equals times the diameter, the diameter of the circle is 18. Therefore AB is a diameter. However, point C is still free to "slide" around the circumference of the circle giving different areas for the triangle, so this is still insufficient to solve for the area of the triangle.

(1) AND (2) SUFFICIENT: Note that inscribed triangles with one side on the diameter of the circle must be right triangles. Because the length of the diameter indicated by Statement (2) indicates that segment AB equals the diameter, triangle ABC must be a right triangle. Now, given Statement (1) we recognize that this is a 30-60-90 degree triangle. Such triangles always have side length ratios of 1: sqrt3 :2

Given a hypotenuse of 18, the other two segments AC and CB must equal 9 and 9sqrt3 respectively. This gives us the base and height lengths needed to calculate the area of the triangle, so this is sufficient to solve the problem.

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For the triangle shown, where A, B and C are all points on  [#permalink]

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Why is B not enough ? If AB is the diameter then triangle ABC has to be 30-60-90 triangle and if that's the case then we can calculate the area right ?

I had read somewhere that if the Hypotenuse is confirmed as the diameter then the triangle is always a right angle triangle.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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Hi anuj11,

With the information in Fact 2, we know that we have a right triangle with a hypotenuse of 18, but we don't know ANYTHING else.

With that info, we have A^2 + B^2 = 18^2, but we don't know the actual values of A and B (which we need to figure out the area). If we had info about the other ANGLES, then we COULD figure those sides out though (since we'd have a relationship among the 3 sides based on the angles). However, we don't know whether the triangle is a 30/60/90 or some other right triangle... and those different options would have different areas. As such, Fact 2 is INSUFFICIENT.

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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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Can any one please explain why B alone is not sufficient.?
Is it because angle A and B could either of measures (i.e. 45,45 or 30,60).?

Posted from my mobile device
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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Can any one please explain why B alone is not sufficient.?
Is it because angle A and B could either of measures (i.e. 45,45 or 30,60).?

Posted from my mobile device

The second statement is not sufficient because even though we know that ABC must be a right triangle we don't know measures of its remaining angles. It's not necessary ABC to be either 30-60-90 or 45-45-90, it can be any right triangle, say 0.5-89.5-90 or say 17-73-90, ...

Hope it helps.
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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Bunuel wrote:
Can any one please explain why B alone is not sufficient.?
Is it because angle A and B could either of measures (i.e. 45,45 or 30,60).?

Posted from my mobile device

The second statement is not sufficient because even though we know that ABC must be a right triangle we don't know measures of its remaining angles. It's not necessary ABC to be either 30-60-90 or 45-45-90, it can be any right triangle, say 0.5-89.5-90 or say 17-73-90, ...

Hope it helps.

THANKS. it did
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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Can any one please explain why B alone is not sufficient.?
Is it because angle A and B could either of measures (i.e. 45,45 or 30,60).?

Posted from my mobile device

even though we know that ABC must be a right triangle we don't know measures of its remaining angles
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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Bunuel wrote:
The second statement is not sufficient because even though we know that ABC must be a right triangle we don't know measures of its remaining angles. It's not necessary ABC to be either 30-60-90 or 45-45-90, it can be any right triangle, say 0.5-89.5-90 or say 17-73-90, ...

Hope it helps.

Bunuel If it were given in the question that angle ACB = 90, then can we conclude that AB is the diameter of the circle?
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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RMD007 wrote:
Bunuel wrote:
The second statement is not sufficient because even though we know that ABC must be a right triangle we don't know measures of its remaining angles. It's not necessary ABC to be either 30-60-90 or 45-45-90, it can be any right triangle, say 0.5-89.5-90 or say 17-73-90, ...

Hope it helps.

Bunuel If it were given in the question that angle ACB = 90, then can we conclude that AB is the diameter of the circle?

Yes.

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.
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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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enigma123 wrote:
Attachment:
Untitled.png
For the triangle shown, where A, B and C are all points on a circle, and line segment AB has length 18, what is the area of triangle ABC?

(1) Angle ABC measures 30°.
(2)The circumference of the circle is $$18\pi$$.

St 1

We don't know anything about the other angles and we cannot assume that this is a 90 degree triangle unless one the sides of the triangle is the diameter of the circle

insuff

St 2

Even if we know that this is a right triangle we do not know whether this is an isosceles right triangle or a scalene right triangle

insuff

St 1 & 2
180- 30 = 60

If we know the remaining angle is 60 then the we can just adjust for ratio 1 \sqrt{3} 2

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Re: For the triangle shown, where A, B and C are all points on  [#permalink]

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Bunuel wrote:
honchos wrote:
I think C is not correct.

A alone is sufficient.

Hypotenuse is given = 18

we know that diameter makes a right angle over the circumference C = 90degree

so from A alone we conclude:

it is 30-60-90 triangle One side is known other two sides can be calculated, Hence area of triangle ABC can be calculated, Do you agree Bunuel?

Let me ask you a question: how do you know that AB is the diameter there? How do you know that the triangle is right angled?

Hi Bunuel,
What is the correct answer as per you?
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For the triangle shown, where A, B and C are all points on  [#permalink]

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In order to find the area of the triangle, we need to find the lengths of a base and its associated height. Our strategy will be to prove that ABC is a right triangle, so that CB will be the base and AC will be its associated height.

(1) INSUFFICIENT: We now know one of the angles of triangle ABC, but this does not provide sufficient information to solve for the missing side lengths.

(2) INSUFFICIENT: Statement (2) says that the circumference of the circle is 18. Since the circumference of a circle equals times the diameter, the diameter of the circle is 18. Therefore AB is a diameter. However, point C is still free to "slide" around the circumference of the circle giving different areas for the triangle, so this is still insufficient to solve for the area of the triangle.

(1) AND (2) SUFFICIENT: Note that inscribed triangles with one side on the diameter of the circle must be right triangles. Because the length of the diameter indicated by Statement (2) indicates that segment AB equals the diameter, triangle ABC must be a right triangle. Now, given Statement (1) we recognize that this is a 30-60-90 degree triangle. Such triangles always have side length ratios of

$$1:\sqrt{3}:2$$

Given a hypotenuse of 18, the other two segments AC and CB must equal 9 and 9 respectively. This gives us the base and height lengths needed to calculate the area of the triangle, so this is sufficient to solve the problem.

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