In the figure above, the radius of circle with center O is 1

Question

https://gmatclub.com/forum/download/file.php?id=66282 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 tmp.JPG Untitled.png

Bunuel · Accepted Answer

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. https://gmatclub.com/forum/download/file.php?id=11920 Hope it helps. Math_Tri_inscribed.png

hogann · Answer

You could use Pythagorean Theorum Here to solve AB.

AC = Diameter or Rx2 = 2
BC = 1

x^2+1^2=2^2 
 x^2=3 
 x=sqrt3

Now we have  (b*h)/2 = (sqrt3*1)/2

B!

zonk · Answer

All triangles inscribed in a circle with the hypotenuse as the diameter of the circle are right triangles.

Therefore

x^2 + 1^2 = 2^2
x = sqrt(3)
Area = x*1*(1/2) = sqrt(3)/2

Aleehsgonji · Answer

Given AC = 2, BC = 1.

Since AC is the diameter and B is a point on the circle, triangle ABC is a right angled triangle.

Area of the triangle = 1/2 * base* height

In the figure, base and height are BC & AB.
AB = \sqrt{3}

So area of the triangle = 1/2 *\sqrt{3}*1 = \sqrt{3}/2
Hence B

bhanushalinikhil · Answer

You should try and avoid assuming any detail in GMAT. Though, you can use a similar way for POE.

Here, the solution is linked to the height of the triangle. 
Given, BC = 1, Radius = 1. ie OB = OC = 1. 
Therefore, we can say that /\ OBC is an equilateral triangle.
Height of equilateral triangle is SQRT(3)/2 * (side)^2.
Therefore, height = SQRT(3)/2.
Therefore, area of /\ ABC = (1/2)*2 (Given) * SQRT(3)/2.
Area = SQRT(3)/2.
Ans : B.

Hope this helps.

abhas59 · Answer

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?

bblast · Answer

I did not apply pythagoras.

I solved this pretty quickly, as I interpreted this as a right triangle inscribed in a circle as 30-60-90 triangle.

so longer leg = 1/2*hypotenuse*root3.

area = 1/2 *root 3 * 1

hope I am correct ?

KarishmaB · Answer

Yes, you are. There are multiple ways of arriving at the value of AB.
You see that Cos C = 1/2 so C must be 60 degrees
 Sin 60 = \sqrt{3}/2  so  AB = \sqrt{3} 
(Let me point out here that you are not expected to know trigonometry in GMAT.)

Or since the sides are 1 and 2, the third side must be  \sqrt{3}  by Pythagorean theorem.

bblast · Answer

thanks karishma

so whenever I see a triangle in a circle and one arm of triangle as diameter if circle, I can quickly solve it using 30-60-90 formulas ?

KarishmaB · Answer

Actually no. Here we know that the diameter is 2 units and one side is 1 unit so the triangle is 30-60-90. It may not always be the case. Say, if the diameter is 2 and one side is given as root 2, it will be a 45-45-90 triangle. 
GMAT generally questions you on one of 30-60-90 and 45-45-90...

bblast · Answer

Thank u . +1 Kudos

But it makes sense that if one triangle-circle holds this relation. Any triangle having the diameter as the hypotenuse should. As the angles for it in a circle will not change. Only the sides will grow or shrink. But ya maybe the ratio 1:\sqrt{3} :2 occurs at this particular size of the 2 geometric figures !!

So I will put ur words in my flashcard.

MrMicrostrip · Answer

These type of question can easily be solved by applying a simple formula.

A question in which triangle is given within a circle and we find out the area of triangle,then we can apply the formula given below

A= abc/4R 
 where a,b,c are the side of a triangle and R is the circum radius.

here, 
 a = AC = 2
 b = BC = 1
 c = AB = √3 ( AB²+BC²=AC² )
 R = 1

Put these values on the formula, we get
 A= (2*1*√3)/(4*1)
 A= √3/2 Ans.

For such formula's, click the link below
 https://gmatclub.com/forum/fundas-of-geometry-part-i-114507.html

KarishmaB · Answer

To help you visualize, I will leave you with a diagram. Ques2.jpg

melguy · Answer

Hi Karishma

I wanted to clarify my misunderstanding. In 45:45:90 the multiples will be x:x:  \sqrt{2x} 
So if the diameter is 2 the other 2 sides should be 2 and 2 \sqrt{2x} ?

Thanks

KarishmaB · Answer

In 45-45-90, the sides will be in the ratio 1:1: \sqrt{2}  or you can say they will be x, x and  \sqrt{2}x  (Mind you, the root is only on 2, not on x).  \sqrt{2}x , the longest side, is the hypotenuse.

Here, the diameter is the hypotenuse i.e. the longest side. The right angle is opposite to the diameter. We know that the side opposite to the largest angle is the longest side. Hence the diameter has to be the longest side i.e.  \sqrt{2}x .
If  \sqrt{2}x  = 2
x =  \sqrt{2}  
So the other two equal sides will be  \sqrt{2}  each.

Zarrolou · Answer

The triangle is a right triangle because one side is the diameter of the circle.

Calculate the third side as  \sqrt{2^2-1^2}=\sqrt{3}

I created a rectangular by translating the triangle (see picture), and the area will be half of the rectangular's. 
 AreaRec=b*h=1*\sqrt{3} 
 AreaTri=\frac{AreaRec}{2}=\sqrt{3}/2 
B

Let me know if this helps

reza52520 · Answer

Guyz why the height is not 1. isn't it obvious that the height is also a radius of the circle !!

Bunuel · Answer

The height is the perpendicular from a vertex to the opposite side. So, both AB and CB are heights as well as the perpendicular from B to AC. Which one turns to be a radius?

BrainLab · Answer

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  
Answer B

In the figure above, the radius of circle with center O is 1

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In the figure above, the radius of circle with center O is 1

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