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Re: quicker method [#permalink]
27 Mar 2005, 11:34
praveen_rao7 wrote:
I am looking for a quicker method ?? If equilateral triangle of area 9Sqrt(3) is inscribed with in a circle what is the area of the circle?
i do not have any quicker methos than this.........
[sqrt(3) L^2]/4= 9sqrt(3), which gives you L=6
draw Prependicular lines from each vertex of the triangle to the other respective sides(base) of the triangle. now we have different triangle. take one triangle and find the H (Hypoteneous) of that triangle and the H is the redius of that triangle.
H of smal triangle = r = 2sqrt(3). then, area of the circle= formula.......
Praveen, if you have any short cut method, pls do share with us.
A triangle if inscribed in a circle is a circumcircle. So lets suppose S is the centre of the circle and A.B and C be the vertices.
Then SA=SB=SC=circumradius
sqrt(3)/4*a^2 = 9*sqrt(3)
a=6 side of triangle
Now angle ASB = 2*Angle C = 2* 60=120(Where S is the Circumcenter . Angle at vertex is half at the circumcenter ,S)
Since AB=6 ===>> AS = 6*Sin120 = 3
So circumradius = 3 and the area of circumcircle = PI*9
PLease ignore the previous posting of mine...
Last edited by gmat2me2 on 27 Mar 2005, 15:48, edited 1 time in total.
Well Given (sqrt3)/4 a^2 = 9sqrt3 => a = 6 length of equilateral triangle
Now The centroid of an equilateral triangle is where the 3 medians cut each other and it should be the centre of the circle and the radius then becomes (2/3)*a*cos 60 deg = 2/3*6*1/2 = 2
Therefore area = pi84 = 4pi
What I would do is to memorize the special right triangle whose three angels are 90, 60, 30. The ratio of its three sides are 1:sqrt(3):2.
The unit area of this special triangle is sqrt(3)/2. Since the area of our triangle (Call it ABC) here is 9sqrt(3), half of the triangle (ABD) is 9sqrt(3)/2. We know that the triangle's three sides very quickly: 3,3sqrt(3),6.
AOD is again this same kind of special triangle and AD=3. We know that AO:AD=2:sqrt(3). So AO=r=6/sqrt(3)=2sqrt(3).
From here it's easy to get the area of the circle. S=pi*r^2=12pi.
Verification: The area of the triangle is 9sqrt(3)=9*1.7=15
Area of the circle =12pi=12*3.14=37
If you draw the graph, you'll see that the area that is formed by AB and the arc AB is a little bit larger than the area of the triangle ABO. It is true for the other two triangles. In other words, the total area of the triangle ABC is a little smaller than the area within the circle and outside of the triangle. In other words it should be a little less than half of the area of the circle. Our results checked out.
If it is a PS question and you have different choices to choose from. You may not even need to calculate if you remember the area of the circle is a little bit larger than twice of the area of the triangle.
Well Given (sqrt3)/4 a^2 = 9sqrt3 => a = 6 length of equilateral triangle
Now The centroid of an equilateral triangle is where the 3 medians cut each other and it should be the centre of the circle and the radius then becomes (2/3)*a*cos 60 deg = 2/3*6*1/2 = 2 Therefore area = pi84 = 4pi
Sorry Anirban you cannot take centroid in all cases... But it is an eq trangle you may probably but it fails for othercases......