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Triangle ABC is inscribed in a semicircle centered at D. What is the

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New post 23 May 2018, 05:52
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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
semicircle.jpg [ 11.36 KiB | Viewed 823 times ]

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

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New post 23 May 2018, 16:53
Bunuel wrote:
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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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Re: Triangle ABC is inscribed in a semicircle centered at D. What is the  [#permalink]

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New post 23 May 2018, 19:53
ADB=120, AD = DB = RADIUS
therefore, AD:DB:AB = 1:1:root(3), which gives us AD=DB=radius=2root(3).
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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New post 28 May 2018, 02:19
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OA : B
Image

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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New post 28 May 2018, 15:31
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
semicircleedit.png [ 64.89 KiB | Viewed 530 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, \(A=Leg_1*Leg_2*\frac{1}{2}\)

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

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


\(A = 6\sqrt{3}\)

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

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New post 06 Feb 2019, 06:41
Bunuel wrote:
Image
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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Re: Triangle ABC is inscribed in a semicircle centered at D. What is the   [#permalink] 06 Feb 2019, 06:41
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