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Re: In the diagram to the right, equilateral triangle ADE is [#permalink]
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pipe19 wrote:
I came across the same problem. I have a question regarding the explanation that is given in the book for how to solve it without estimating.

They say that if we draw a line from the center of the circle O to D it bisects the angle of the triangle and creates another 30:60:90 triangle.

Why does the line OD bisect the angle? Is there some kind of rule when a line bisects an angle?

Thankful for any suggestions.


Refer diagram below

If I get you correct, you are talking about the red lines drawn bisecting the 60 deg angle to 30 deg

Yes ; that's correct because the CG of inscribed circle = CG of equilateral triangle

The red line drawn is joining the midpoint of the opposite side; so it will bisect; That's the rule
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Re: In the diagram to the right, equilateral triangle ADE is [#permalink]
PareshGmat wrote:
pipe19 wrote:
I came across the same problem. I have a question regarding the explanation that is given in the book for how to solve it without estimating.

They say that if we draw a line from the center of the circle O to D it bisects the angle of the triangle and creates another 30:60:90 triangle.

Why does the line OD bisect the angle? Is there some kind of rule when a line bisects an angle?

Thankful for any suggestions.


Refer diagram below

If I get you correct, you are talking about the red lines drawn bisecting the 60 deg angle to 30 deg

Yes ; that's correct because the CG of inscribed circle = CG of equilateral triangle

The red line drawn is joining the midpoint of the opposite side; so it will bisect; That's the rule


Yes, exactly! So because the center of the circle = center of triangle the line bisects the angle?
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Re: In the diagram to the right, equilateral triangle ADE is [#permalink]
pipe19 wrote:
PareshGmat wrote:
pipe19 wrote:
I came across the same problem. I have a question regarding the explanation that is given in the book for how to solve it without estimating.

They say that if we draw a line from the center of the circle O to D it bisects the angle of the triangle and creates another 30:60:90 triangle.

Why does the line OD bisect the angle? Is there some kind of rule when a line bisects an angle?

Thankful for any suggestions.


Refer diagram below

If I get you correct, you are talking about the red lines drawn bisecting the 60 deg angle to 30 deg

Yes ; that's correct because the CG of inscribed circle = CG of equilateral triangle

The red line drawn is joining the midpoint of the opposite side; so it will bisect; That's the rule


Yes, exactly! So because the center of the circle = center of triangle the line bisects the angle?


For this problem, we can summarize as follows

1. The inscribe circle touches the midpoints of all sides of the equilateral triangle
2. The lines in red are the medians of the equilateral triangle
3. Medians also bisect the angle 60 of equilateral triangle to angle 30
4. In an equilateral triangle, the median, angle bisector, and altitude are equal. They intersect at center of the inscribed and circumscribed circle.
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Re: In the diagram to the right, equilateral triangle ADE is [#permalink]
mahendru1992 wrote:
5. The median is divided in the ratio of 2:1 (someone confirm please)


Agree.. This is correct
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In the diagram to the right, equilateral triangle ADE is [#permalink]
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Say r=1 and O the center of the circle. Then because of ADE=60, OD biscepts ADE to 30+30.
Therefore we have a triangle ODF with OF=1 and FD= √3 . (OF is the line that biscepts ED to its half).
We know that 30-60-90 triangle have sides 1-√3 - 2.

Ac=πr^2=π*1=π
Asq=ΕD^2= (2 √3 ) ^2 = 12

Ac/Asq=π/12
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Re: In the diagram to the right, equilateral triangle ADE is [#permalink]
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mahendru1992 wrote:
Attachment:
Untitled.png
In the diagram to the right, equilateral triangle ADE is drawn inside square BCDE. A circle is then inscribed inside triangle ADE. What is the ratio of the area of the circle to the area of the square?

(A) \(\frac{\pi}{12}\)

(B) \(\frac{\pi}{8}\)

(C) \(\frac{\pi}{6}\)

(D) \(\frac{\pi}{4}\)

(E) \(\frac{\pi}{2}\)

Let the side of the square is X, and AP is the perpendicular distance from vertex A to side ED.
Side of the equilateral triangle is also X according to the figure
Half of an equilateral triangle(AEP) is 30-60-90 triangle, and sides are in the ratio of a:a√3:2a
a=X/2(half of the side of the equilateral triangle)
AP=√3X/2
Since it is inscribed circle, centriod and the center of the circle will be same. AP will be median, and centroid will divide it in 2:1(2Y:1Y)
3Y=AP=√3X/2
Y=X/2√3(Radius of the circle)
area of the circle/area of the square=pi*Y^2/X^2
=(pi*X^2/12)/X^2
=pi/12
A:)
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Re: In the diagram to the right, equilateral triangle ADE is [#permalink]
Rule 1: the In-radius of an Inscribed Circle inside an Equilateral Triangle = (1/3) * (Height of Equilateral Triangle)

Since the question asks for a Ratio answer, we can choose Numbers to fit the Diagram.

Let Side of Equilateral Triangle = 1

The Side of the Equilateral Triangle on the Bottom shares the Side of the Square ------> thus Side of Square = 1

Area of Square = (1)^2 = 1


In-radius of Inscribed Circle:

Height of the Equilateral Triangle is given by = (side) * sqrt(3) * (1/2)

(1/3) * (Height of Equilateral Triangle) = Radius of Inscribed Circle

given side =1:

(1/3) * (1) * sqrt(3) * (1/2) = sqrt(3)/6 = Radius of Inscribed Circle

Area of Inscribed Circle = (pi) * (r)^2 = (sqrt(3) / 6)^2 * (pi) = 3(pi) / 36 = (pi)/12


Ratio of the: (Area of Circle) / (Area of Square) = [ (pi) / 12 ] / [1] = (pi) /12

-A-
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Re: In the diagram to the right, equilateral triangle ADE is [#permalink]
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Re: In the diagram to the right, equilateral triangle ADE is [#permalink]
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