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Triangle BDC is an equilateral triangle and triangle ABC is a right tr

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Triangle BDC is an equilateral triangle and triangle ABC is a right tr  [#permalink]

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New post 04 Sep 2018, 23:06
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Triangle BDC is an equilateral triangle and triangle ABC is a right triangle. If line segment DC is 6 units in length, what is the area of the shaded region?


A. 9

B. \(9\sqrt{3}\)

C. 18

D. \(18\sqrt{2}\)

E. \(18\sqrt{3}\)


Attachment:
image023.jpg
image023.jpg [ 2.71 KiB | Viewed 323 times ]

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Triangle BDC is an equilateral triangle and triangle ABC is a right tr  [#permalink]

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New post 05 Sep 2018, 00:17
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Bunuel wrote:
Image
Triangle BDC is an equilateral triangle and triangle ABC is a right triangle. If line segment DC is 6 units in length, what is the area of the shaded region?


A. 9

B. \(9\sqrt{3}\)

C. 18

D. \(18\sqrt{2}\)

E. \(18\sqrt{3}\)


Attachment:
image023.jpg


ADB is an isosceles triangle with AD=BD
Since DBC is an equilateral triangle, so DB=BC=DC=6
So, AD=6

So, AC=12. Hence using Pythagorean formula in the right angled triangle ABC, we have \(AB=AC^2-BC^2\)
Or, \(AB=\sqrt{12^2-6^2}=6\sqrt{3}\)
So, area of triangle ABC=\(\frac{1}{2}*6*6\sqrt{3}=18\sqrt{3}\)

Area of equilateral triangle DBC=\(\sqrt{3}/4*6^2\)=\(9\sqrt{3}\)

Area of shaded region=Area of triangle ABC-Area of triangle DBC=\(18\sqrt{3}-9\sqrt{3}=9\sqrt{3}\)

Ans. (B)
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Triangle BDC is an equilateral triangle and triangle ABC is a right tr &nbs [#permalink] 05 Sep 2018, 00:17
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