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Math Expert V
Joined: 02 Sep 2009
Posts: 58398
Triangle ABC is inscribed in a semicircle centered at D. What is the  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 59% (02:32) correct 41% (02:53) wrong based on 64 sessions

HideShow timer Statistics Triangle ABC is inscribed in a semicircle centered at D. What is the area of Triangle ABC?

A. $$\frac{12}{\sqrt{3}}$$

B. $$6 \sqrt{3}$$

C. $$12$$

D. $$12 \sqrt{3}$$

E. $$18 \sqrt{3}$$

Attachment: semicircle.jpg [ 11.36 KiB | Viewed 957 times ]

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GMAT 1: 540 Q49 V16 GMAT 2: 680 Q49 V33 Triangle ABC is inscribed in a semicircle centered at D. What is the  [#permalink]

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Bunuel wrote: Triangle ABC is inscribed in a semicircle centered at D. What is the area of Triangle ABC?

A. $$\frac{12}{\sqrt{3}}$$
B. $$6 \sqrt{3}$$
C. $$12$$
D. $$12 \sqrt{3}$$
E. $$18 \sqrt{3}$$
Attachment:
semicircle.jpg

According to the figure & Q-stem : DB =DC ==> angle DBC =angle DCB & as angle BDC is 60, angle DBC =angle DCB=60
Now the triangle ABC is inscribed in a semicircle in such a way that AC is the diameter. Thus angle B is 90 . Hence angle A=30
Hence tan 30 = BC/6 ==> BC =6 *tan30 = 6*$$\frac{1}{\sqrt{3}}$$= $$\frac{6}{\sqrt{3}}$$
Hence area of ABC= $$\frac{1}{2}$$ *AB*BC= $$\frac{1}{2}$$ * $$6$$*$$\frac{6}{\sqrt{3}}$$=$$6 \sqrt{3}$$

Hence I would go for option B.
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Manager  B
Joined: 16 Apr 2018
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Re: Triangle ABC is inscribed in a semicircle centered at D. What is the  [#permalink]

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BDC=60, and DB=DC=radius, therefore all angles are 60.
Altitude (root(3)/2)*side. Side = radius = 2root(3).

Area of triangle = 1/2*base*height=1/2*(2*2root(3))*((root(3)/2)*2root(3))=6root(3)

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Re: Triangle ABC is inscribed in a semicircle centered at D. What is the  [#permalink]

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1
1
OA : B As In $$\triangle$$ BDC,$$\angle$$BDC = 60$$^{\circ}$$, BD=DC=r as D is the centre of semicircle.
$$\angle$$DCB =$$\angle$$DBC
$$\angle$$BDC +$$\angle$$DCB +$$\angle$$DBC = 180$$^{\circ}$$
60$$^{\circ}$$+2$$\angle$$DCB =180$$^{\circ}$$
$$\angle$$DCB=$$\angle$$DBC = 60$$^{\circ}$$

So $$\triangle$$ BDC is equilateral triangle , with all Sides, i.e $$BD=BC=DC=r$$

Now applying pythagoras theorem in $$\triangle$$ ABC , we get
$$AB^2+BC^2=AC^2$$
$$6^2+r^2=(2r)^2$$
$$36 = 4r^2-r^2$$
$$r^2=\frac{36}{3}=12$$
$$r=\sqrt{12}$$

Area of $$\triangle$$ ABC = $$\frac{1}{2}*BC*AB$$
=$$\frac{1}{2}*\sqrt{12}*6=3\sqrt{12}=6\sqrt{3}$$
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Re: Triangle ABC is inscribed in a semicircle centered at D. What is the  [#permalink]

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Bunuel wrote:
Triangle ABC is inscribed in a semicircle centered at D. What is the area of Triangle ABC?

A. $$\frac{12}{\sqrt{3}}$$

B. $$6 \sqrt{3}$$

C. $$12$$

D. $$12 \sqrt{3}$$

E. $$18 \sqrt{3}$$

Attachment: semicircleedit.png [ 64.89 KiB | Viewed 654 times ]

An inscribed triangle whose hypotenuse is the diameter
is a right triangle.

Hypotenuse AC = diameter
Δ ABC is a right triangle

$$\sqrt{3}$$ in the answer choices
should give notice: this right triangle is probably 30-60-90

Δ ABC IS a 30-60-90 triangle

• It has one 90° angle (∠ ABC)

• It has one 30° angle, ∠ BAC, DERIVED:

∠ ADB = 120° : it lies on a straight line with given ∠ BDC = 60°
Straight line property: (180 - 60) = 120°

Bisect the 120° angle: Draw a line from D to X

Now there are two small 30-60-90 right triangles
-- At bisected 120° angle are two 60° angles
-- DX is perpendicular to AB = two 90° angles at X
-- the third angles of both triangles must = 30°

That is, ∠ BAC = 30°, which is one of Δ ABC 's angles

• Third angle in right Δ ABC must = 60°
That is, ∠ BCA = 60° (interior angles = 180)

Find Leg Lengths from ratios and given side AB = 6

Right Δ ABC is a 30-60-90 triangle with sides
opposite corresponding angles in the ratio

$$x : x\sqrt{3} : 2x$$

AB = $$6$$, opposite the 60° angle, corresponds with $$x\sqrt{3}$$

• Side BC = $$x$$
$$6 = x\sqrt{3}$$

BC = $$x = \frac{6}{\sqrt{3}}$$

Area?

Area, $$A=Leg_1*Leg_2*\frac{1}{2}$$

$$A=(6*\frac{6}{\sqrt{3}}*\frac{1}{2})=\frac{18}{\sqrt{3}}$$

$$A=(\frac{18}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}})=\frac{18\sqrt{3}}{3}=6\sqrt{3}$$

$$A = 6\sqrt{3}$$

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Re: Triangle ABC is inscribed in a semicircle centered at D. What is the  [#permalink]

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Bunuel wrote: Triangle ABC is inscribed in a semicircle centered at D. What is the area of Triangle ABC?

A. $$\frac{12}{\sqrt{3}}$$

B. $$6 \sqrt{3}$$

C. $$12$$

D. $$12 \sqrt{3}$$

E. $$18 \sqrt{3}$$

Attachment:
semicircle.jpg

So nothing special here, 30 60 90 triangle 1:root3:2

Apart from that this is a triangle in a circle which has makes a 90 degree angle at ABC

We have one angle 60 this means Triangle BDC is an equilateral triangle with all angles 60 degree

<BAC = 30 degree

30 60 90 triangle 1:$$\sqrt{3}$$:2

We follow this ratio, this makes BC = 2 $$\sqrt{3}$$

Now = 1/2 * 2 $$\sqrt{3}$$* 6

$$6 \sqrt{3}$$
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