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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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over2u wrote:
Attachment:
sample.JPG
In the figure above, the radius of the circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

(A) √2/2
(B) √3/2
(C) 1
(D) √2
(E) √3

My way
Spoiler: :: Description
We should find the area of triangle with this formula: 0.5*h*b. It is quite easy but we don't know the length of heigh. So what I do in this situation: I just imagine that h equal BC, so the area of triangle is 2*0.5*1=1, but in fact its smaller, because heigh is slightly less than BC. So we can look on answers and define the aproriate value for our triangle area. It is definetly less than 1, sqr(2), sqr(3). So we have two choises: sqr(2)/2 which equals to 1.4/2 or 0.7 and sqr(3)/2 which equals to 1.7/2 or 0.85. I pick the biggest because heigh just SLIGHTLY less then BC. So the answer is B

To solve this one you don't even need to use pythagoras formula. It's a right triangle (bcs. Hypotenus=Diameter of the circle) with a hypotenuse of 2 and one side equal to 1. Here you have a 90-60-30 triangle, hence the third side is equal $$\sqrt{3}$$ --> $$Area = \sqrt{3}*1/2$$
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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Always remember: A circle inscribed in a semicircle with always be a right angle triangle.
In this case,
∠ABC = 90

Hence in the triangle ABC, we have base = BC = 1
Hypotenuse = CA = 2
Height = AB^2 = 2^2 + 1^2
Hence AB = √3

Therefore, area = 1/2*√3*1 = √3/2

Correct option: B
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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I have two ways for this:

According to theory since one of triangle's side is the diameter the triangle is a right triangle. In case you have any reservations about the hypotenuse Recall the rule which states: The larger side is opposite to to the larger angle. Clearly the larger side AC hence the angle B has to be 90 degrees. BUT this also means that since the larger side is AC is also the hypotenuse.

By using Pythagoras theorem AC^2 = BC^2 + AB^2 and given the info from the question you have 2^2 - 1 = AB^2 hence AB = sqrt(3). Also AB is the HEIGHT of the right triangle and thats why we bother to find it. Thus area = (base * height ) / 2 = (1 * sqrt(3) ) / 2

Another method

By drawing the MEDIAN BO we divide the triangle into two triangles with EQUAL areas (by definition). The MEDIAN BO splits the side AC into two EQUAL length lines hence AO = 1 and OC = 1 (By definition). Therefore the triangle BCO is a equilateral triangle with side r and the are of an equilateral is given by ( r^2 * sqrt(3) ) / 4 = ( 1 * sqrt(3) ) / 4

According to the MEDIAN properties the MEDIAN divides the triangle into TWO triangle with EQUAL areas hence the total are of ABC = 2*Area(BCO) hence sqrt(3) / 2 .

I think the second method is faster....
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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How do you know that BC is the base and AC is the height? Is it because on right angle triangle the 90 degree angle is never the base or the height? Bunuel
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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1
rishit1080 wrote:
How do you know that BC is the base and AC is the height? Is it because on right angle triangle the 90 degree angle is never the base or the height? Bunuel

The altitude (height) of a triangle is the perpendicular from the base to the opposite vertex (the base may need to be extended). Since there are three possible bases, there are also three possible altitudes. In case of a right triangle both legs are altitudes as well as perpendicular from right angle to hypotenuse.
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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Bunuel wrote:
abhas59 wrote:
In the figure above, the radius of circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

Hi guys,
Just one doubt, how can we say that triangle ABC is a right angled triangle given the radius is one & other side BC is one.
I mean to say that is there any property for circle that if radius is equal to any one side of the triangle, then the inscribed triangle so formed will be right-angled one?

We know that AC is a diameter. There is a property of a right triangle inscribed in circle:

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 side, then that triangle is a right triangle and diameter is hypotenuse.
Attachment:
Math_Tri_inscribed.png

Hope it helps.

Usually it's always a GMAT rule that you should just assume ANYTHING...

The question only states that "the radius of the circle with center O is 1 and BC = 1".

How do I know that AC is the diameter? Yes, I see it on the graph but the graphs aren't accurate, are they? A could also be placed 1mm before the circle's line. It's not stated anywhere that the triangle is inscribed, is it?

Thanks for explaining!
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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guenthermat wrote:
Bunuel wrote:
abhas59 wrote:
In the figure above, the radius of circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

Hi guys,
Just one doubt, how can we say that triangle ABC is a right angled triangle given the radius is one & other side BC is one.
I mean to say that is there any property for circle that if radius is equal to any one side of the triangle, then the inscribed triangle so formed will be right-angled one?

We know that AC is a diameter. There is a property of a right triangle inscribed in circle:

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 side, then that triangle is a right triangle and diameter is hypotenuse.
Attachment:
Math_Tri_inscribed.png

Hope it helps.

Usually it's always a GMAT rule that you should just assume ANYTHING...

The question only states that "the radius of the circle with center O is 1 and BC = 1".

How do I know that AC is the diameter? Yes, I see it on the graph but the graphs aren't accurate, are they? A could also be placed 1mm before the circle's line. It's not stated anywhere that the triangle is inscribed, is it?

Thanks for explaining!

First of all, if it were the way you are saying, how would you solve the question? It's a PS question, so one of the answers should be correct. How you get it?

Next, GMAT will not use such tricks. Check what is OG saying about the figures.

Problem Solving
Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Data Sufficiency:
Figures:
• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2).
• Lines shown as straight are straight, and lines that appear jagged are also straight.
• The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero.
• All figures lie in a plane unless otherwise indicated.
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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Hi all,

Why can't AC be the base, and a vertical line drawn down from B, the height? I keep getting answer A, Sqrt(2)/(2).

The vertical line from B forms a smaller 45-45-90 triangle to the right, with BC as the hypotenuse, and the height of the entire triangle becomes 1/sqrt(2). Since the base is then 2 times radius, or 2, the Area = b*h/2, so the answer is A. Where am I going wrong?
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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TippingPoint93 wrote:
Hi all,

Why can't AC be the base, and a vertical line drawn down from B, the height? I keep getting answer A, Sqrt(2)/(2).

The vertical line from B forms a smaller 45-45-90 triangle to the right, with BC as the hypotenuse, and the height of the entire triangle becomes 1/sqrt(2). Since the base is then 2 times radius, or 2, the Area = b*h/2, so the answer is A. Where am I going wrong?

A triangle has three bases, so yes you can consider AC to be the base. The problem is that the perpendicular from B to AC does not create 45-45-90 triangle. Check the image below: Attachment: triangle1.png [ 8.55 KiB | Viewed 1586 times ]

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In the figure above, the radius of circle with center O is 1  [#permalink]

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A video explanation can be found here:

Three of the many equations and facts you'll need to memorize for the GMAT are

- Area of a triangle = 1/2(base)(height)

- In a right triangle with angles 30-60-90 degrees, if the leg opposite the 30 degree angle has length a, then the hypotenuse has a length of 2a, and the other leg has a length of a√3 (see video below)

- If a triangle inscribed in a circle has the diameter of the circle as one of its sides, that side will always be the hypotenuse, and the two legs that touch the circumference of the crcle will always interesect at 90 degrees (also best explained in video below)

Here we have hypotenuse = 2, therefore base = 1 and height = √3, therefore area = 1/2(1)(√3) = √3/2

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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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Hi, Bunuel,

I solved this question by applying the formula for a right triangle S = 1/2 a*b, where a and b are the legs of triangle.

I also wanted to solve it via the method of height and base, but could not get the correct answer. Please could you help to understand how to find height here if we forget that triangle ABC is a right triangle?

Thanks. Re: In the figure above, the radius of circle with center O is 1   [#permalink] 23 Jul 2019, 11:09

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