In the diagram above, ABC is an equilateral triangle. D is the midpoi

Question

cpotg_img2.png In the diagram above, ABC is an equilateral triangle. D is the midpoint of AC. BD is a diameter of the circle. If AD = 4, what is the area of the circle? A. 8\sqrt{3}\pi B. 12\pi C. 16\pi D. 16\sqrt{3}\pi E. 48\pi Kudos for a correct solution .

ynaikavde · Answer

BD =  \sqrt{AB^2- AD^2} =4 \sqrt{3} 
radius=2 \sqrt{3} 
area=r^2* pi =12 pi

answer is B

iamdp · Answer

Hi

Considering triangle BDC

DC= 4 and BC= 8 
Using Pythagorean BD = 4√3

No radius of the circle = BD/B = 2√'3
Therefore area = pi r^2 = 12pi

kasiaw99 · Answer

I think the answer is B.

Applying the pythagorean theorom and the 30-60-90 rule we know that BD is 4√3. The get the radius we divide by 2 to get 2√3.
To get the area of the circule we apply the formula: A=r^2pi.
A=(2√3)^2pi
A=12pi

peachfuzz · Answer

Using pythagorean theorem, the height of the equilateral triangle is sqrt(48), which is also the diameter of the circle. In this case the radius will be equal to sqrt(12). 
The area of the circle will then be 12*pi

Answer: B

23a2012 · Answer

the answer is B
the hight is the diameter of the circle so the hight =x sq.r3/2 where x is the linght of the tringle side
the AD in half AC so AC as basic will be equal to 2(AD)=8 
so BD=8*sq.r3/ and the raduies equal to half the BD=4sq.r3
the area =pi (4sq.r)^2= pi12

gmat6nplus1 · Answer

height of an equilateral triangle = s√3/2 where s is the measure of the triangle's side.

s=8; height=4√3=diameter; radius=2√3; Area=12(3,14)

answer B.

rahultherockpandey · Answer

AD=4, so AC=8 (as D was midpoint)...now the diameter BD is the height of the triangle...so BD=((sq.rt3)/2)*8=4sqrt3.
now radius=1/2*BD=2sqrt3.....so area of the circle is pie*(2sqrt3)^2=12pie

PareshGmat · Answer

Answer = B. 12\pi cpotg_img2.png Area of equilateral triangle = \frac{\sqrt{3}}{4} * 8^2 = 16\sqrt{3} Let the diameter of circle = height of triangle = h Again, area = \frac{1}{2} * 8 * h = 16\sqrt{3} h = 4\sqrt{3} Area of circle = \pi (2\sqrt{3})^2 = 12\pi

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION:

Because ABC is an equilateral triangle, right triangle ABD is a 30-60-90 triangle, one of the  GMAT favorite triangles . From the ratios of that triangle:  \frac{BD}{AD}=\frac{\sqrt{3}}{1}  -->  BD=4\sqrt{3} .

Half of that is radius of the circle:  r=2\sqrt{3}

Use that to find the area:  A=\pi{r^2}=\pi{(2\sqrt{3})^2}=12\pi

Answer = (B)

bhatiavai · Answer

Since ABC is an Equilateral triangle and BD is an altitude from B to AC, It will bisect AC. AD=DC ==> AC = 8

Area of Equilateral triangle = √3/4 * AC^2 = 16√3

Area of Triangle = 1/2 * base * Height = 16√3 ==> 1/2 * AC * BD ==> BD = 4√3

BD is the Diameter of the circle hence radius of circle = 2√3
Area of Circle = (pi) r^2 == > 12pi

Ans B

jhaamod10 · Answer

AD:BD=1:√3
Hence BD=4√3
Radius=2√3
Area of circle = π(2√3)^2=12π
Ana is B

Posted from my mobile device

bumpbot · Answer

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In the diagram above, ABC is an equilateral triangle. D is the midpoi

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