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Let "\(a\)" be the length of the side of the Equilateral Triangle

The Radius of a circle circumscribing am Equilateral triangle, \(R = a/\sqrt{{3}}\)

Hence \(a = 4\sqrt{{3}}\)

In an Equilateral triangle the Altitude is also the median. hence D is midpoint of AC.

So we have AD = \(2\sqrt{{3}}\)

Join A to O & form right triangle AOD.

\({AO}^2 = {OD}^2 + {AD}^2\)

we get \({OD}^2 = 4^2 - {(2\sqrt{{3}})}^2\)

\(OD = 2\)

Hence we have \(DE = 4 - 2 = 2\).


Answer C.



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Bunuel
In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?

A. 1
B. \(\sqrt{3}\)
C. 2
D. \(2\sqrt{3}\)
E. \(4\sqrt{3}\)
Attachment:
eIdABed 2018.07.09.jpg
eIdABed 2018.07.09.jpg [ 33.01 KiB | Viewed 9147 times ]
I. \(r = \frac{2}{3} h\)

BD is an altitude (= median -- equilateral triangle)
The medians of an equilateral triangle intersect (= centroid) at the center of the circumscribed circle

The centroid divides the median into a ratio of \(2x:1x\)
Median = Altitude, \(h = 3x\)
The radius of the circle is \(\frac{2}{3}\) the altitude of the triangle
\(r = \frac{2}{3}h\)
\(h =\frac{3}{2}r\)

BE is the diameter of the circle = \(2r = 8\)
Height, BD, =\(\frac{3}{2} r =(\frac{3}{2}*4)= 6\)

(BE - BD) = DE
\((8 - 6) = 2\)
\(DE = 2\)

Answer C

II. 30-60-90 triangles
The medians (= altitudes) of an equilateral triangle divide it into six congruent 30-60-90 triangles.
∆ OAD is one of those triangles
-- BD is an altitude
-- equilateral triangle: altitude = median = perpendicular bisector of vertex and base

The ratio of sides opposite 30-60-90 angles is \(x : x\sqrt{3}: 2x\)
OD corresponds with \(x\)
AD corresponds with \(x\sqrt{3}\)
OA corresponds with \(2x\)

OA is a radius. OA = \(4\)
OA, opposite the 90° angle = \(2x\)
OA is twice the length (\(2x:x\)) of OD.

OD = \(2\)
OE = \(4\) (radius)

The length of DE is just (OE - OD) = \((4 - 2) = 2\)

\(DE = 2\)

Answer C
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Here, in the picture, I provide an appropriate explanation to the question. Using the properties of the equilateral triangle, we can find the way to solve this problem easily.
Attachments

abcde.PNG
abcde.PNG [ 80.38 KiB | Viewed 8835 times ]

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DE = BE - BD
BE = diameter =2r= 2*4=8

Solved with the help of only two properties:

1. Side of an equilateral triangle inscribed in a circle is r√3
r=4
So, side= 4√3

2. The altitude of an equilateral triangle is √3/2a, where a= side of the equilateral triangle.
So, BD (altitude) = √3/2*4√3 = 6

Therefore, DE = 8 - 6 = 2. ANS
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OE = 4 (radius)

Solve for 30-60-90 triangle OAD
x:xsqrt(3):2x

2x=4
x=2 --> length of OD

DE = OE-OD = 4-2 = 2
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The Radius (Circumradius) of an Inscribed Equilateral Triangle inside a Circle = 2/3rd of Altitude

Logic: Median = Altitude = Angle Bisector from every Vertex in an Equilateral Triangle. Further, the Centroid (Intersection of 3 Medians) divides the Altitude of an Equilateral Triangle into a Ratio of 2 : 1, where the Circumradius drawn from the Centroid to the Vertex of the Triangle = 2/3 of the Median (which is the same as the Altitude)

4 = (2/3) * h ; where h = Altitude BD = Median BD

h = 6 = BD
BE = Diameter = 8

BE - BD = DE, or 8 - 6 = 2 Answer Choice C
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Bunuel

In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?

A. 1
B. \(\sqrt{3}\)
C. 2
D. \(2\sqrt{3}\)
E. \(4\sqrt{3}\)


Solution:

Attachment:
Isosceles in Circle.png
Isosceles in Circle.png [ 15.03 KiB | Viewed 6638 times ]

Let’s connect points O and C with a line segment, creating triangle ODC, which is a 30-60-90 triangle, because if we let the angle BCO be y, then the angle CBO is also y since BOC is an isosceles triangle. The angle DOC is 2y (since it is the sum of BCO and BOC) and the angle OCD is 90 - 2y. Now, since angle BCD is 60 degrees, we have y + (90 - 2y) = 60 and thus, y = 30. It follows that OCD = 60 - 30 = 30 and the triangle ODC is a 30-60-90 right triangle. The triangle ODC has side ratios of x : 2x : x√3. We know that the hypotenuse OC is 4, so side OD has length 2. We know that OE has length 4, so the length of DE is 2.

Answer: C
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Bunuel

In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?

A. 1
B. \(\sqrt{3}\)
C. 2
D. \(2\sqrt{3}\)
E. \(4\sqrt{3}\)


Attachment:
eIdAB.png

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/07 ... relations/

We know that side of the inscribed equilateral triangle = sqrt(3) * Radius of the circle

Since radius is 4, side = 4*sqrt(3)

Altitude of an equilateral triangle = \(\frac{\sqrt{3}}{2} * Side = \frac{\sqrt{3}}{2} * 4*\sqrt{3} = 6\)

BE, the diameter of the circle is 8 and BD (altitude of triangle) is 6.

So DE = BE - BD = 8 - 6 = 2

Answer (C)
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