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

Question

https://gmatclub.com/forum/download/file.php?id=41391 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} semicircle.jpg

u1983 · Answer

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 .

thinkTheta · Answer

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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Princ · Answer

OA : B
 https://gmatclub.com/forum/download/file.php?id=41391

As In  	riangle  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  	riangle  BDC is equilateral triangle , with all Sides, i.e  BD=BC=DC=r

Now applying pythagoras theorem in  	riangle  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  	riangle  ABC =  \frac{1}{2}*BC*AB 
 = \frac{1}{2}*\sqrt{12}*6=3\sqrt{12}=6\sqrt{3}

generis · Answer

semicircleedit.png 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

KanishkM · Answer

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}

bumpbot · Answer

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

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