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 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 = 6Right 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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