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Re: M20 #07 : TRIANGLE INSIDE A CIRCLE [#permalink]

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30 Apr 2010, 13:10

ykpgal wrote:

S = (BC∙AC)/2

BC = 1/2∙AB, => AB² = BC² + AC² => AB² = 1/4∙AB² + AC² AC = (√3)/2∙AB => S = ((1/2∙AB)∙(√3/2∙AB))/2

S = (√3/8)AB²

Um, no idea what you did here. I see S = the area, but it looks like you didn't find a value for the area.

I also don't see how some people have mentioned that the base of the triangle lies along the diameter. All we know is that the triangle is inscribed inside the circle.

I got C. I don't know how to find the area, but I know it's possible.

St 1 tells us it's a right triangle, since the square of one side, equals the sum of the squares of the other two sides. But as there could be many different sizes of right traingles that would fit insiade this circle, this statement is insufficient.

St2 gives us another one of the angles. Like statement one, alone it is insufficent, but together we know know this is a 30-60-90 triangle. This will fit into this circle only one way. I don't care that I can't figure out the area, only that I know it's possible.

Re: M20 #07 : TRIANGLE INSIDE A CIRCLE [#permalink]

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04 May 2010, 13:59

nifoui wrote:

gmatbull wrote:

(1) implies a Pythagorean triple of diameter (a side) 2 units. Height and other angles are unknown, so Insufficient.

(2) we can't find area of a triangle with just an angle given. Insufficient

(1) and (2): implies a triangle with 30(1) : 60(sqr{3} : 90(2) angle - sides combination. So, Sufficient.

How do we know that statement 1 implies a pythagorean theorem? I don't understand the formula in statement 1...

Help?

The formula in statement 1 says the length of side AB squared equals the the sum of the squares of the other two sides (BC and AC). This is the pythagorean theorem.

When I first looked at the equation, I was thinking \(A*B^2\) but then I realized what kind of equation I was looking at and knew AB was the length of the side from A to B. Seems like others saw the same thing at first.

S2 says angle CAB is 30 degree. Can't we find other angles of the triangle and apply 1:\sqrt{3}:2 equatio to find sides.

The sides of a triangle are always in the ratio \(1:\sqrt{3}:2\) when we have 30°-60°-90° right triangle but knowing that just one angle of the inscribed triangle equals to 30° does not mean that we have this case.

Points A, B, and C lie on a circle of radius 1. What is the area of triangle ABC?

(1) \(AB^2 = BC^2 + AC^2\) --> triangle ABC is a right triangle with AB as hypotenuse --> \(area=\frac{BC*AC}{2}\). Now, 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). So, hypotenuse AB=diameter=2*radius=2, but just knowing the length of the hypotenuse is not enough to calculate the legs of a right triangle thus we can not get the area. Not sufficient.

(1)+(2) From (1) ABC is a right triangle and from (2) \(\angle CAB=30\) --> we have 30°-60°-90° right triangle and as AB=hypotenuse=2 then the legs equal to 1 and \(\sqrt{3}\) --> \(area=\frac{BC*AC}{2}=\frac{\sqrt{3}}{2}\). Sufficient.

Statement 1: Tells us ABC is a right triangle. Property of a right triangle inscribed within a circle- The hypotenuse of the triangle is necessarily the diameter of the circle the triangle is inscribed within. So we know the hypotenuse=2. However, 2 find the area, we need to know the other 2 sides as well and with this statement alone, we have no means to find those. So, this statement alone is insufficient.

Statement 2: Tells us one angle is 30 degrees. This information alone cannot lead us to find anything at all.

(1) and (2) combined: Tells us that the triangle is a 30-60-90 one. From (1), we know that the biggest side is 2. In a 30-60-90 triangle, the sides are in the ratio of 1:\sqrt{3}:2, where 1 represents smallest side. So, the legs of the triangle will be 1 and \sqrt{3} and thats all the information we need to find out the area. You can go ahead and solve the problem but since its a DS question, that is unnecessary.

Answer: C
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I dont think that statement A is sufficient. As C can be anywhere on the circle. So we do not get the hight every time we connect O to C which means the area of the triangle can not be determined by 1 alone.

St 2 is also not sufficient.

But if we combine 1 and 2, we get all the angles of triangle which are 30, 60 and 90 degrees. and the length of the hypotenuse by which we can calculate other sides as well. AB = 2 AC = sqrt(3) and BC = 1.

so the area = (1/2)*Base*Height = (1/2)*1*sqrt(3) = sqrt(3)/2

so my answer is together the statements are sufficient.
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Re: M20 #07 : TRIANGLE INSIDE A CIRCLE [#permalink]

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05 May 2011, 07:53

havok wrote:

Is statement one an equation that people memorized? I don't see the 90 degree angle reference.

Thats Pythagoras Theorem for you. It States that In a 90 degree triangle, the square of the longest side is equal to the sum of the squares of the other two sides. This is an exclusive property of 90 degree triangles.

So, if you happen to see a triangle with sides AB, BC and AC and if it is mentioned that \(AB^2= BC^2+AC^2\), you can be sure that its a 90 degree triangle and that AB is the longest side (the hypotenuse)

Problems based on Pythagoras theorem are extremely common on the GMAT.

Follow this link for more: math-triangles-87197.html There is a topic on Right Triangle in this chapter.
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Re: M20 #07 : TRIANGLE INSIDE A CIRCLE [#permalink]

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11 May 2011, 20:46

Hi Folks, This is my first post and i would like to take this opportunity to answer this question.

Answer is C. Reason - a) AB is diameter = 2. Angle ACB = 90. but ration od AC &AB not clear. Insufficient.

b) Angle CAB is 30 degree. Insufficient.

a) +b) value of AC and AB can be calculated because CAB is 30 degree. and since it is right angle triangle - Area will be (AC+CB)/2.

Answer C.

Cheers, Aj.
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Why does statement 1 necessary convey the idea of a right triangle.....In one of the gmat club test questions I saw a property -

For triangles which has all acute angles there are, two cases where this property is true

such as AB^2 = BC^2 + AC^2 and BC^2 = AB^2 + AB^2

For triangle with an obtuse angle, AB^2 < BC^2 + AC ^2

For right angled triangle, there is only one case of AB ^2 = BC^2 + AC^2 (as opposed to two cases as found in acute angled triangle)

Then why will statement A necessarily suggest a right angled triangle, it can also be an a triangle with all acute angles.

Where did you see that? It's not true.

Must know for the GMAT: converse of the Pythagorean theorem is also true. For any triangle with sides a, b, c, if a^2 + b^2 = c^2, then the angle between a and b measures 90°.

Generally if c is the longest of the three sides of a triangle then: If a^2 + b^2 = c^2, then the triangle is right. If a^2 + b^2 > c^2, then the triangle is acute. If a^2 + b^2 < c^2, then the triangle is obtuse.