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

Question

https://gmatclub.com/forum/download/file.php?id=43275 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} image023.jpg

BrentGMATPrepNow · Accepted Answer

Since ∆BDC is an equilateral triangle, and since ∆ABC is a right triangle, we can add the following to our diagram. 
 https://i.imgur.com/GI1jHC5.png

Next, if we add the following perpendicular line . . . 
 https://i.imgur.com/MUU1Opa.png 
. . . we can add more to our diagram

And we can add this.
 https://i.imgur.com/QptotUL.png

At this point, we can simplify matters A LOT by recognizing that if we "cut" out the top triangle and rotate it . . . 
 https://i.imgur.com/af8jPwe.png

And keep rotating . . . 
 https://i.imgur.com/Jp81dO5.png 
. . . 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 
 https://i.imgur.com/67QURXO.png

Area of an equilateral triangle = (√3)(side²/4)
So, the area of the shaded triangle = (√3)( 6 ²/4)
= (√3)(36/4)
= 9√3

Answer: B

Cheers,
Brent

PKN · Answer

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)

ScottTargetTestPrep · Answer

Solution: triangle regions.png From the diagram above, we see that regions 1 and 2 have the same area as regions I and II, respectively. Therefore, the area of the shaded region (triangle ABD) is the same as the area of the unshaded region (triangle BDC), which is an equilateral triangle. Recall that the area of an equilateral triangle with side s is s^2√3/4, therefore, the area of triangle BDC is 6^2√3/4 = 36√3/4 = 9√3 and this is also the area of triangle ABD, i.e., the shaded region. Answer: B

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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

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