In figure below, equilateral triangle ABC is inscribed in circle O, wh

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.



Thanks,
GyM

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

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.

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

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

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

Solution:


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

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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