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

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Math Expert
Joined: 02 Sep 2009
Posts: 49303
Triangle BDC is an equilateral triangle and triangle ABC is a right tr  [#permalink]

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05 Sep 2018, 00:06
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00:00

Difficulty:

35% (medium)

Question Stats:

80% (02:17) correct 20% (02:13) wrong based on 15 sessions

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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 [ 2.71 KiB | Viewed 211 times ]

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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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05 Sep 2018, 01:17
1
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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Triangle BDC is an equilateral triangle and triangle ABC is a right tr &nbs [#permalink] 05 Sep 2018, 01:17
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# Triangle BDC is an equilateral triangle and triangle ABC is a right tr

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