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

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New post 16 Jan 2019, 17:32
Top Contributor
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


Since ∆BDC is an equilateral triangle, and since ∆ABC is a right triangle, we can add the following to our diagram.
Image

Next, if we add the following perpendicular line . . .
Image
. . . we can add more to our diagram

And we can add this.
Image

At this point, we can simplify matters A LOT by recognizing that if we "cut" out the top triangle and rotate it . . .
Image

And keep rotating . . .
Image
. . . it fits nicely here.

At this point, we should see that the shaded area is actually a nifty EQUILATERAL triangle with sides of length 6
Image

Area of an equilateral triangle = (√3)(side²/4)
So, the area of the shaded triangle = (√3)(6²/4)
= (√3)(36/4)
= 9√3

Answer: B

Cheers,
Brent
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Re: Triangle BDC is an equilateral triangle and triangle ABC is a right tr   [#permalink] 16 Jan 2019, 17:32
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