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Comparison of Area of Triangles inside a Circle [#permalink]
05 Oct 2016, 06:05
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Question Stats:
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AB is the Diameter and point O is the center of the circle shown above
Quantity A 
Quantity B 
Area of Triangle ABC 
Twice the Area of Triangle OBD 
A : Quantity A is greater B : Quantity B is greater C : Both quantities are equal D : Answer Cannot be determined




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Re: Comparison of Area of Triangles inside a Circle [#permalink]
05 Oct 2016, 12:20
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Hey, Let radius of the triangle be r. Now since AB is the diameter of the circle, triangle ABC is a right angle triangle. And from triangle properties we know value of angle \(x = 30\) degrees. Thus triangle OBD is a equilateral triangle. Now AC= \(2*r*cos(x)\)=\(2*r*cos(30)\)=\(2*r*\sqrt{3}*\frac{1}{2}\)=\(\sqrt{3}r\). And BC= \(2*r*sin(x)\)=\(2*r*sine(30)\)=\(2*r*\frac{1}{2}\)=\(r\) Area of traingle ABC= \(\frac{1}{2}\)*AC*BC=\(\frac{1}{2}\sqrt{3}r^2\). Now Area of the equilateral triangle OBD= \(\frac{1}{4}\sqrt{3}r^2\). Clearly 2* Area of OBD= Area of ABC. Hence C is the correct option.
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Re: Comparison of Area of Triangles inside a Circle [#permalink]
09 Oct 2016, 01:20
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I personally recommend to avoid using Sine,Cosine and other trigonometric function to solve geometry related problems as other easier ways seem more handy and quick.Here is my solution: <BCA is an inscribe angle and AB is the diameter,so <BCA = 90 degree.So as ABC is a right triangle so we can use Pythagorean Theorem on this triangle.Since sum of the interior angles of any triangle is equal to 180 degrees therefore m<BCA+x+2x = 180 => 90+3x = 180 => x = 30 => Triangle ABC is a 306090 triangle and we may use properties of 306090 triangle to solve this question.Sides of a 306090 triangle are in the ratio 1:\(sqrt3\):2.Since hypotenuse of Triangle ABC is equal to Diameter AB = 2r of the circle;where r is radius of the circle.So the sides of Triangle ABc are r,r\sqrt{3} and 2r. => Quantity A : Area of Triangle ABC = (1/2)*(base)*(height) = (1/2)*(BC)*(AC) = = (1/2)*(r)*(r\(sqrt3\)) = \((sqrt3/2)*r^2\) Since 2x = 2*30 = 60 so Triangle OBD is an equilateral triangle and formula for area of an equilateral triangle is \((sqrt3/4)*s^2\),where s is the side of the equilateral triangle.Side of equilateral Triangle OBD is equal to radius r of the circle Quantity B : Twice the Area of Triangle ABC = \(2*(sqrt3/4)*r^2\) = \((sqrt3/2)*r^2\) which is equal to Quantity => Quantity A = Quantity B.Hence answer is C. This question has been taken from tests of Greatest Prep.Here is the link to the website of Greatest Prep : http://www.greatestprep.com



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Re: Comparison of Area of Triangles inside a Circle [#permalink]
15 Nov 2018, 07:24
yasir9909 wrote: I personally recommend to avoid using Sine,Cosine and other trigonometric function to solve geometry related problems as other easier ways seem more handy and quick.Here is my solution: <BCA is an inscribe angle and AB is the diameter,so <BCA = 90 degree.So as ABC is a right triangle so we can use Pythagorean Theorem on this triangle.Since sum of the interior angles of any triangle is equal to 180 degrees therefore m<BCA+x+2x = 180 => 90+3x = 180 => x = 30 => Triangle ABC is a 306090 triangle and we may use properties of 306090 triangle to solve this question.Sides of a 306090 triangle are in the ratio 1:\(sqrt3\):2.Since hypotenuse of Triangle ABC is equal to Diameter AB = 2r of the circle;where r is radius of the circle.So the sides of Triangle ABc are r,r\sqrt{3} and 2r. => Quantity A : Area of Triangle ABC = (1/2)*(base)*(height) = (1/2)*(BC)*(AC) = = (1/2)*(r)*(r\(sqrt3\)) = \((sqrt3/2)*r^2\) Since 2x = 2*30 = 60 so Triangle OBD is an equilateral triangle and formula for area of an equilateral triangle is \((sqrt3/4)*s^2\),where s is the side of the equilateral triangle.Side of equilateral Triangle OBD is equal to radius r of the circle Quantity B : Twice the Area of Triangle ABC = \(2*(sqrt3/4)*r^2\) = \((sqrt3/2)*r^2\) which is equal to Quantity => Quantity A = Quantity B.Hence answer is C. This question has been taken from tests of Greatest Prep.Here is the link to the website of Greatest Prep : http://www.greatestprep.comhow is it possible if 2x=60, the triangle is equilateral. how do we calculate angle ODB and DBO ? Please, answer.



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Re: Comparison of Area of Triangles inside a Circle [#permalink]
15 Nov 2018, 08:37
AE wrote: yasir9909 wrote: I personally recommend to avoid using Sine,Cosine and other trigonometric function to solve geometry related problems as other easier ways seem more handy and quick.Here is my solution: <BCA is an inscribe angle and AB is the diameter,so <BCA = 90 degree.So as ABC is a right triangle so we can use Pythagorean Theorem on this triangle.Since sum of the interior angles of any triangle is equal to 180 degrees therefore m<BCA+x+2x = 180 => 90+3x = 180 => x = 30 => Triangle ABC is a 306090 triangle and we may use properties of 306090 triangle to solve this question.Sides of a 306090 triangle are in the ratio 1:\(sqrt3\):2.Since hypotenuse of Triangle ABC is equal to Diameter AB = 2r of the circle;where r is radius of the circle.So the sides of Triangle ABc are r,r\sqrt{3} and 2r. => Quantity A : Area of Triangle ABC = (1/2)*(base)*(height) = (1/2)*(BC)*(AC) = = (1/2)*(r)*(r\(sqrt3\)) = \((sqrt3/2)*r^2\) Since 2x = 2*30 = 60 so Triangle OBD is an equilateral triangle and formula for area of an equilateral triangle is \((sqrt3/4)*s^2\),where s is the side of the equilateral triangle.Side of equilateral Triangle OBD is equal to radius r of the circle Quantity B : Twice the Area of Triangle ABC = \(2*(sqrt3/4)*r^2\) = \((sqrt3/2)*r^2\) which is equal to Quantity => Quantity A = Quantity B.Hence answer is C. This question has been taken from tests of Greatest Prep.Here is the link to the website of Greatest Prep : http://www.greatestprep.comhow is it possible if 2x=60, the triangle is equilateral. how do we calculate angle ODB and DBO ? Please, answer. In ∆OBD, O is 2x=60.. Now OB=OD =radius, so angle OBD =angle ODB OBD+ODB=18060=120, so both are 120/2=60, as both are equal
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Re: Comparison of Area of Triangles inside a Circle [#permalink]
20 Nov 2018, 07:54
yasir9909 wrote: AB is the Diameter and point O is the center of the circle shown above
Quantity A 
Quantity B 
Area of Triangle ABC 
Twice the Area of Triangle OBD 
A : Quantity A is greater B : Quantity B is greater C : Both quantities are equal D : Answer Cannot be determined For a triangle where the side is the diameter the opposing angle will be 90 degree. Given that AB's angle is 90. Then we have 90 + x + 2x = 180 so 3x= 90 and x = 30. Given that 2x = 60 And OB = OD (both are radius) then they have the same corresponding angle. As a result it becomes an equilateral triangle. The area of an equilateral = (s^2 *sqrt of 3)/4 https://www.mathopenref.com/triangleequ ... larea.htmlWhere the side is the radius. For ABC the triangle is 30:60:90 so it has the following values s:s sqrt of 3:2s https://www.themathpage.com/aTrig/306090triangle.htmThe area of ABC = (s * s sqrt of 3)/2 = (s^2 * sqrt of 3)/2 2 times the area of equilateral equals = 2 * (s^2 *sqrt of 3)/4 = (s^2 *sqrt of 3)/2 Hence both are equal.



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Re: Comparison of Area of Triangles inside a Circle [#permalink]
20 Nov 2018, 20:36
Answer: A (I need to double check, Not sure about the answer)
AB is the diameter Point O is the center of the circle A: Area of triangle ABC? B: twice the Area of triangle OBD? Consider point C, the environment of circle which it faces (I think you name it arc) equals 180, why? Because AB is diameter and AB arc equals 180, so it’s facing angle which is C equals half of it, so C is 90 degrees. A: Now area of ABC = AC*BC/2 We have: A+B+C = 180 degrees x+2x+90 = 180 > x = 30degrees So angle O is 2x=60degrees. And as OB and OD are radiuses of circle, the OBD triangle is a equilateral. And thus, angles B and D equal (1802x)/2 = 60degrees B: The area of OBD*2 = 2* BD*H/2= BD*H How much is H? we have all angles in OBD sin B = H/OB > H = sinB * OB = sin60 * OB = RADICAL3/2 *OB A = AC*BC/2 sinB(2x)= AC/AB > √3/2 = AC/AB > AC = √3/2*AB > A= √3/2*AB*BC/2 B = √3/2 * OB A= √3/2*AB*BC/2 B = √3/2 * OB So we omit √3/2 from both. A > AB*BC/2 B > OB We know OB = AB/2, so we substitute it in B: A > AB*BC/2 B > AB/2 We omit AB/2 from both And we will have A > BC B > 1 BC is bigger than 1, so A is bigger than B. (Not sure about the answer)
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Re: Comparison of Area of Triangles inside a Circle [#permalink]
20 Nov 2018, 20:42
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FatemehAsgarinejad wrote: Answer: A (I need to double check, Not sure about the answer) AB is the diameter Point O is the center of the circle A: Area of triangle ABC? B: twice the Area of triangle OBD?
Consider point C, the environment of circle which it faces (I think you name it arc) equals 180, why? Because AB is diameter and AB arc equals 180, so it’s facing angle which is C equals half of it, so C is 90 degrees. A: Now area of ABC = AC*BC/2
We have: A+B+C = 180 degrees x+2x+90 = 180 > x = 30degrees So angle O is 2x=60degrees. And as OB and OD are radiuses of circle, the OBD triangle is a equilateral. And thus, angles B and D equal (1802x)/2 = 60degrees
B: The area of OBD*2 = 2* BD*H/2= BD*H
How much is H? we have all angles in OBD sin B = H/OB > H = sinB * OB = sin60 * OB = RADICAL3/2 *OB
A = AC*BC/2 sinB(2x)= AC/AB > √3/2 = AC/AB > AC = √3/2*AB > A= √3/2*AB*BC/2 B = √3/2 * OB
A= √3/2*AB*BC/2 B = √3/2 * OB So we omit √3/2 from both. A > AB*BC/2 B > OB
We know OB = AB/2, so we substitute it in B: A > AB*BC/2 B > AB/2 We omit AB/2 from both And we will have A > BC B > 1 BC is bigger than 1, so A is bigger than B. (Not sure about the answer)
you have forgotten to put BD back in the calculations as shown in coloured portion. so finally A > BC B > BD and BC=BD as they both are equal to the radius.. therefore A=B answer is C
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Re: Comparison of Area of Triangles inside a Circle [#permalink]
01 Dec 2018, 16:07
Can someone please explain this?



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Re: Comparison of Area of Triangles inside a Circle [#permalink]
01 Dec 2018, 20:24
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kruttikaaggarwal wrote: Can someone please explain this? Take triangle ABC.. As the hypotenuse is diameter, the triangle is right angled at C. Therefore x+2x=90....3x=90....x=30 So ABC is 306090 triangle with sides 1:√3:2 or a:√3a:2a... The hypotenuse AB is diameter and equal to 2r, so 2r=2a...a=r So area of ABC = 1/2 * AC * BC =r*√3r/2=√3r^2/2 Now let's see OBD.. Angle at centre is 2x=2*30=60.. The other two angles will be equal as the sides are equal to radius.. Therefore the other two angles are (18060)/2=60 each. So OBD is an equilateral triangle with sides r, so area = (√3/4)*r^2= (1/2)(√3/2*r^2)=(1/2) area of ABC.. So A=B C
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Some useful Theory. 1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressionsarithmeticgeometricandharmonic11574.html#p27048 2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effectsofarithmeticoperationsonfractions11573.html?sid=d570445335a783891cd4d48a17db9825 3. Remainders : https://greprepclub.com/forum/remainderswhatyoushouldknow11524.html 4. Number properties : https://greprepclub.com/forum/numberpropertyallyourequire11518.html 5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolutemodulusabetterunderstanding11281.html



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Re: Comparison of Area of Triangles inside a Circle [#permalink]
08 Nov 2019, 16:21
chetan2u wrote: kruttikaaggarwal wrote: So OBD is an equilateral triangle with sides r, so area = (√3/4)*r^2= (1/2)(√3/2*r^2)=(1/2) area of ABC.. I get that (1/2)(√3/2*r^2)=(1/2) will be half of the area of ABC but how do we then establish that the area of the equilateral and the 306090 are the same?




Re: Comparison of Area of Triangles inside a Circle
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