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by fibbonnaci » Thu Jan 28, 2010 12:17 pm
in an equilateral triangle, the centroid, orthocentre, circumcentre etc all coincide.

well known property is that centroid divides the median in the ratio 2:1

if i connect the center of the circle O to the three vertices of the triangle say OA, OB, OC- then,
OA= OB=OC = 2/3 of the median (according to the centroid division ratio)

OA, OB and OC are infact the radius of the circle = 4

therefore say OA = 2/3 of median = 4.

median= 6.

In equilateral triangle median is the height of the triangle.

height of equilateral triangle is root (3)/2 * side of the triangle= 6
therefore side of the triangle is 6*2 / root(3)

perimeter = 3 * (12/ root (3) ) => 12 root(3)

Hope this helps!
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by sars72 » Thu Jan 28, 2010 12:23 pm
I just know the forumula:
For a circle circumscribing an equilateral triangle, as is the case here: R = [a * (3)^(1/2)]/3
where R is the radius of the circle and a is the length of one side of the equilateral triangle

Rewriting the equation -> a = 3R/[(3)^(1/2)]

substituting R with 4 -> a = 12/[(3)^(1/2)]

Remember that we need the perimiter -> 3 * 12/[(3)^(1/2)]

-> 36/[(3)^(1/2)]

multiplying both numerator & denominator by [(3)^(1/2)] -> [36(3)^(1/2)]/3

-> 12(3)^(1/2)
tada!
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by money9111 » Thu Jan 28, 2010 2:25 pm
do we not use anything re: 30 - 60 - 90 triangle where the ratio is:

X, X sqrt(3), 2X?
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by ajith » Thu Jan 28, 2010 4:04 pm
money9111 wrote:do we not use anything re: 30 - 60 - 90 triangle where the ratio is:

X, X sqrt(3), 2X?
To use 30 - 60 - 90 triangle property we have to find out such a triangle first

The Origin, One Vertex, Center of the chord is such a triangle

For simplicity let me call the center O, vertex A and center of the chord E

Angle OAC =30; AOE is 60 and OEA is 90, further OA is 4 (radius)

now OE =2x =4; AE =2Sqrt(3)

AB = 2*AE = 4sqrt(3); perimeter is 3AB =12sqrt(3)
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by money9111 » Thu Jan 28, 2010 4:40 pm
thanks makes sense
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by sanju09 » Fri Jan 29, 2010 2:57 am
skorolkova wrote:An equilateral triangle is inscribed within a circle with radius = 4.
What is the perimeter of the triangle?
Image
In such a case, the radius of circle, say r, happens to be 2/3 of one median of the inscribed equilateral triangle of side a. For the length of one median of this inscribed equilateral triangle of side a, we can quickly form a 30-60-90 triangle with a as the side opposite to the right angle, and hence a √3/2 will be each median, whose 2/3 is the radius of circle.

Over to question now, if a is each side of the inscribed equilateral triangle, then


a √3/2*(2/3) = 4, or a = 12/√3

Perimeter of the inscribed equilateral triangle of side 12/√3 is 36/√3 or [spoiler]12 √3[/spoiler].
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by harshavardhanc » Fri Jan 29, 2010 6:42 am
phew!!! formulae!... tough to remember them...


the good thing about this problem is it has equilateral triangle... symmetry everywhere!!!!


solution...

Let the side of the triangle be X and the center of the figure be O. (see it's not only the center of the circle but of the triangle as well)

from any vertex of the triangle join O and drop a perpendicular on the side of the triangle including that vertex.


now when you join a vertex to the center, the line segment bisects 60 degrees ( the angle of equi tria carrying that vertex) . The perpendicular bisects 120 degrees and the opposite side of the triangle, and you've got a 30-60-90 triangle with hypotenuse as the radius and base as X/2.

now simple Pythagorean theorem :

X/2 = Radius * Cos 30

X = 2 * 4 * sqrt(3) / 2

X = 4 * sqrt(3)


Hence, perimeter = 3 * 4 * sqrt(3) = 12sqrt(3)
Regards,
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by tata » Fri Jan 29, 2010 12:00 pm
Money9111, Ajith,

I am using the same approach for 30-60-90 ( 1-SQRT3-2) triangle for which the new ratio is (a-2*4-c) Where c is the side of the equilateral triangle. Using this the side of the triangle comes out to be 16*SQRT(3) and perimiter is 3*16*SQRT(3)
I guess I am doing something wrong, Ajith can you please explain with the figure. The way I figured is draw a perpendicular from one vertex to a side of the triangle, this will make a 30-60-90 triangle, with side opposite 60 degree angle is the diameter, 8 in this case, and side opposite 90 degree angle is side of equilateral triangle.




ajith wrote:
money9111 wrote:do we not use anything re: 30 - 60 - 90 triangle where the ratio is:

X, X sqrt(3), 2X?
To use 30 - 60 - 90 triangle property we have to find out such a triangle first

The Origin, One Vertex, Center of the chord is such a triangle

For simplicity let me call the center O, vertex A and center of the chord E

Angle OAC =30; AOE is 60 and OEA is 90, further OA is 4 (radius)

now OE =2x =4; AE =2Sqrt(3)

AB = 2*AE = 4sqrt(3); perimeter is 3AB =12sqrt(3)
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by ajith » Fri Jan 29, 2010 12:12 pm
tata wrote:Money9111, Ajith,

I am using the same approach for 30-60-90 ( 1-SQRT3-2) triangle for which the new ratio is (a-2*4-c) Where c is the side of the equilateral triangle. Using this the side of the triangle comes out to be 16*SQRT(3) and perimiter is 3*16*SQRT(3)
I guess I am doing something wrong, Ajith can you please explain with the figure. The way I figured is draw a perpendicular from one vertex to a side of the triangle, this will make a 30-60-90 triangle, with side opposite 60 degree angle is the diameter, 8 in this case, and side opposite 90 degree angle is side of equilateral triangle.

Image

In the image given 30 60 90 triangle is BOF and OF:FB:OB =1:Sqrt(3):2

Since we know OB=4 ; FB =4/2*sqrt(3) = 2Sqrt(3) and FB is half of BC BC = 4Sqrt(3) and perimeter 12Sqrt(3)

Your contention that BEC is 30-60-90 triangle is indeed true but you missed out on the measurement of BE
BE is not equal to 2*4, BC is less than that (See fibbonnaci's post to know how to correct the mistake)
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