Carcass wrote:
The measure of one angle in an isosceles triangle is 120°. If the length of the side opposite this angle equals \(12 \sqrt{3}\) , then what is the perimeter of the triangle?
A. \(18 + 12 \sqrt{3}\)
B. \(21 + 12 \sqrt{3}\)
C. \(24 + 12 \sqrt{3}\)
D. \(30 + 12 \sqrt{3}\)
E. \(36 + 12 \sqrt{3}\)
Let ABC be that isosceles triangle where BC = \(12 \sqrt{3}\) and Angle BAC = 120°
Angle ABC + Angle BAC + Angle ACB = 180
2(Angle ABC) + 120° = 180°
Angle ABC = Angle ACB = 30°
Drop a Perpendicular from A to BC and name the point as D, such that BD = DC = \(6 \sqrt{3}\)
Angle ABC = 30°
Angle BAC = 60°
Angle BDA = 90°
In 30°-60°-90° triangle, the ratio of the sides is \(x : \sqrt{3}x : 2x \)
\(\sqrt{3}x = 6 \sqrt{3}\)
\(x = 6\)
Therefore,
BD = DC = 6
AB = AC = 12
Perimeter = \(12 + 12 + 12 \sqrt{3} = 24 + 12 \sqrt{3}\)
Hence, option C