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

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In figure below, equilateral triangle ABC is inscribed in circle O, wh  [#permalink]

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New post 08 Jul 2018, 20:31
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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
eIdAB.png [ 54.87 KiB | Viewed 889 times ]

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In figure below, equilateral triangle ABC is inscribed in circle O, wh  [#permalink]

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New post 08 Jul 2018, 21:15
length of DE = length of BE - length of BD

length of BE = \(2*radius\) = 8

If the radius of a circle is r then the length of the side of the equilateral triangle inscribed in that circle will be \(r\sqrt{3}\)

so length of AB = \(4\sqrt{3}\)

Now triangle ABD is right angled triangle right-angled at D and angle A = 60 -> angle ABD = 30

in 30-60-90 triangle the lengths of sides will be in the ratio \(1:\sqrt{3}:2\)

AD:BD:AB = \(1:\sqrt{3}:2\) and we know AB = \(4\sqrt{3}\)

AD:BD:AB = \(1:\sqrt{3}:2\) = \(2:6:4\sqrt{3}\)

BD = 6

DE = BE - BD = 8-6 = 2

Hence option C
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Re: In figure below, equilateral triangle ABC is inscribed in circle O, wh  [#permalink]

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New post 09 Jul 2018, 02:07
Bunuel wrote:
Image
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


We have to calculate DE, DE=radius of circle,OE-side of the \(\triangle\)OAD,OD
Given radius,r=4
so, side of equilateral \(\triangle\)ABC,BC=\(\sqrt{3}*4\)
In the right angled \(\triangle\)OAD, OA=4,\(AD=\frac{BC}{2}\)=\(2\sqrt{3}\)
So, OD=\(\sqrt{OA^2-AD^2}\)=2
so DE=OE-AD=4-2=2 unit

Ans. (C)
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Re: In figure below, equilateral triangle ABC is inscribed in circle O, wh  [#permalink]

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New post 09 Jul 2018, 02:52
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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In figure below, equilateral triangle ABC is inscribed in circle O, wh  [#permalink]

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New post 09 Jul 2018, 11:03
1
Bunuel wrote:
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 681 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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Re: In figure below, equilateral triangle ABC is inscribed in circle O, wh  [#permalink]

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New post 17 Jul 2018, 07:02
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 528 times ]


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Re: In figure below, equilateral triangle ABC is inscribed in circle O, wh  [#permalink]

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New post 17 Jul 2018, 08:16
1
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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Re: In figure below, equilateral triangle ABC is inscribed in circle O, wh &nbs [#permalink] 17 Jul 2018, 08:16
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