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If a triangle inscribed in a circle has area 40, what is the

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If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 22 Mar 2011, 08:46
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A
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C
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If a triangle inscribed in a circle has area 40, what is the area of the circle?

(1) The base of the triangle is equal to the diameter of the circle.
(2) The measure of one of the angles in the triangle is 30.

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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 22 Mar 2011, 09:08
2
rxs0005 wrote:
If a triangle inscribed in a circle has area 40, what is the area of the circle?

(1) The base of the triangle is equal to the diameter of the circle.
(2) The measure of one of the angles in the triangle is 30.


(1)
This statement tells us that the inscribed triangle is a right angled triangle.
Right angled triangle have three sides;
Two smaller sides and one hypotenuse.

Area = 1/2*(smaller side)*(bigger side)=40 if we assume that it is not an isosceles right triangle.

We know the radius = 1/2*(Diameter) = 1/2*(Hypotenuse) [Hypotenuse is referred as "BASE" in this statement]
Not Sufficient.

(2)
The measure of one of the angles in the triangle is 30.
We can make plenty such triangles with one angle as 30.
Not sufficient.

Combining both;

We know that the triangle is a right angled triangle with 30-90-60 as its angles.

The opposite sides will be in ratio: \(x:2x:\sqrt{3}x\)

\(Area = \frac{1}{2}*x*\sqrt{3}x=40\)
\(x^2=\frac{80}{\sqrt{3}}\)
\(x=\sqrt{\frac{80}{\sqrt{3}}}\)

\(Hypotenuse = Diameter = 2x= 2*\sqrt{\frac{80}{\sqrt{3}}}\)
\(Area = \pi*(\frac{Diameter}{2})^2=\pi*(\sqrt{\frac{80}{\sqrt{3}}})^2=\pi*\frac{80}{\sqrt{3}}\)

Sufficient.

Ans: "C"
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 22 Mar 2011, 18:05
(1) is not sufficient, as the height can be altered, and the area = 1/2 * b * h = 40, where the base can also vary and hence the area of circle will also vary, as base of triangle = diameter.

(2) If measure of one angle is 30, still we can't decide on the other sides (or angles), so not sufficient. FOr example, the base may or may not be the diameter.

(1) and (2) together, If the base is the Diameter, and one angle is 30 degrees, the triangle is a 30.60-90 right triangle with sides in the ratio

1:sqrt(3):2. Hypotenuse is the Diameter and it can be found as area of triangle = 40.

So Answer - C.
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 22 Mar 2011, 20:50
From one we know its a right angle triangle . To calculate area of circle we need to know the radius of the circle(i.e hypotenuse of a triangle (b*h=80) ).

From 2 : its 30:60 :90 triangle with ration 1: 3^1/2 :2
we know b*h=80 or x*3^1/2 x=80.. we get x here

hypotenuse =diameter of circle=2x ...once we know diameter we can calculate area of circle.

Thus c
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 23 Mar 2011, 00:38
It’s C:
The answer is area of a circle, A=πr^2 =π( √(80/(√3)) )^2 here the radius is √(80/(√3))
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 28 Sep 2015, 07:44
i have a doubt in this question.

if we consider diameter as the base of the triangle, then opposite angle will be 90. in this case, two sides making the 90 degree angle will be base and height or vice versa. and in this case, diameter is the longest side. this diameter can not be the base as mentioned in the statement 1. hence E should be the option.

please explain where i am making mistake. i find statement 1 erroneous.
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 08 Aug 2017, 00:14
rxs0005 wrote:
If a triangle inscribed in a circle has area 40, what is the area of the circle?

(1) The base of the triangle is equal to the diameter of the circle.
(2) The measure of one of the angles in the triangle is 30.


Given : Area of triangle say ABC with AC as the longest side is 40
DS: Area of circle.

Statement 1 : AC is the diameter of the circle.. So, /_B = 90 deg. And hence 1/2* AB* BC = 40 i.e. AB * BC = 80
NOT SUFFICIENT to find AC /2

Statement 2: Since measure of one angle is 30 . But we don't know about other angles and hence we can't relate it with the area of triangle to find its sides.
NOT SUFFICIENT

Combined : Third angle is 60.
AB*BC = 80 .........................(i)
Also AB/Sin30 = BC/Sin60 or AB/Sin60 = BC/Sin 30 (ii)
By solving (i) and (ii) for Either case we will get value of AB and BC..

Both cases will finally result in same value of AC = \(\sqrt{{AB^2+BC^2}}\)
Then we can find area of circle = pi*AC^2/4

Answer C
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 31 Aug 2017, 01:20
Ans is C
1) the base is dia of circle but sides and angles are unknown = insufficient .......A and D gone
2) any one angle is 30 which is still insufficient because we dont know sides of triangle, other 2 angles, so cant find r .....B is gone

1+2
base is dia so 1 angle auto equals 90 , other is 30 remaining is 60
3 angles are known, sides of triangle can be calculated in terms of r
No need to find area since this is. DS ques
directly it is sufficient to find r and so our area Hence C is answer
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If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 31 Aug 2017, 02:07
I thought the answer could be A,
(r*2r)/2=40
r^2 =40
r= \sqrt{40}

I really think I am right..
Bunuel or anyone, Please correct me, thank you
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 31 Aug 2017, 02:19
pclawong wrote:
I thought the answer could be A,
(r*2r)/2=40
r^2 =40
r= \sqrt{40}

I really think I am right..
Bunuel or anyone, Please correct me, thank you


You assume that the inscribed triangle is a right isosceles one but we don't know that. All we know that it's a right triangle (it may or may not be isosceles).
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 07 Sep 2017, 23:25
rxs0005 wrote:
If a triangle inscribed in a circle has area 40, what is the area of the circle?

(1) The base of the triangle is equal to the diameter of the circle.
(2) The measure of one of the angles in the triangle is 30.


St 1

Base of triangle= diameter tells us that we have a 90 degree triangle but without any further info we cannot calculate (i.e is this right isosceles or scalene?)

Insuff

St 2

By itself we know nothing for sure about the other two angles

Insuff

St 1 & 2

Using both statements we can deduce that this is a 30-60-90 triangle and simple use

3x= 80

solve for x square etc

C
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 10 Apr 2018, 06:38
I did get this correct but I have a question -- how do we know that the triangle is right once we know that the base is the diameter of the circle. Is that simply a known property of inscribed triangles?
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 10 Apr 2018, 06:44
lostnumber wrote:
I did get this correct but I have a question -- how do we know that the triangle is right once we know that the base is the diameter of the circle. Is that simply a known property of inscribed triangles?


Yes, it's a known property.

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.
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 10 Apr 2018, 07:43
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This property is extension of theorem:An angle at the circumference of a circle is half the angle at the centre subtended by the same arc.ie \(\angle\)APB =\(\frac{1}{2}\)\(\angle\)AOB

Arc(Semicircle) created by Diameter subtends \(180^{\circ}\) at centre, so angle at the circumference subtended by same arc will be \(90^{\circ}\)
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Re: If a triangle inscribed in a circle has area 40, what is the  [#permalink]

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New post 10 Apr 2018, 08:12
Princ wrote:
lostnumber

This property is extension of theorem:An angle at the circumference of a circle is half the angle at the centre subtended by the same arc.ie \(\angle\)APB =\(\frac{1}{2}\)\(\angle\)AOB

Arc(Semicircle) created by Diameter subtends \(180^{\circ}\) at centre, so angle at the circumference subtended by same arc will be \(90^{\circ}\)


This is very useful, thank you for the insight Princ! And thank you as well Bunuel
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