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

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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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

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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