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# A circle is inscribed in equilateral triangle ABC so that

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29 Sep 2004, 03:49
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26. A circle is inscribed in equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on the circle and on line segment AB. If line segment AB = 6, what is the area of the figure created by line segments AD, AE and minor arc DE?

OA 3(3)^1/2

Last edited by saurya_s on 29 Sep 2004, 05:39, edited 1 time in total.
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29 Sep 2004, 05:14
Height= AM = sqrt3/2 x side = 3sqrt3

radius= 1/3 * height = sqrt3

Area of shaded region = sqrt3/4 x (side)^2 - (pi)r^2

= 6.15

We want area of ADE = 6.15/3 = 2.05.
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29 Sep 2004, 07:09
Yep, I learnt this once, but almost forgot. Thanks venksune for bringing this back.

for an inscribed triangle (circle inside triangle), r = h/3
for a circumscribed triangle (triangle inside circle), r = 2h/3

Ans should be 3*sqrt(3) - pi
29 Sep 2004, 07:09
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# A circle is inscribed in equilateral triangle ABC so that

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