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Math Expert V
Joined: 02 Sep 2009
Posts: 57244
Triangle BDC is an equilateral triangle and triangle ABC is a right tr  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 70% (02:24) correct 30% (02:26) wrong based on 68 sessions

### HideShow timer Statistics 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 [ 2.71 KiB | Viewed 947 times ]

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Status: Learning stage
Joined: 01 Oct 2017
Posts: 1032
WE: Supply Chain Management (Energy and Utilities)
Triangle BDC is an equilateral triangle and triangle ABC is a right tr  [#permalink]

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Bunuel wrote: 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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PKN

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Joined: 12 Sep 2015
Posts: 3916
Location: Canada
Re: Triangle BDC is an equilateral triangle and triangle ABC is a right tr  [#permalink]

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Top Contributor
Bunuel wrote: 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. Next, if we add the following perpendicular line . . . . . . we can add more to our diagram

And we can add this. At this point, we can simplify matters A LOT by recognizing that if we "cut" out the top triangle and rotate it . . . And keep rotating . . . . . . 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 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
_________________ Re: Triangle BDC is an equilateral triangle and triangle ABC is a right tr   [#permalink] 16 Jan 2019, 18:32
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