Bunuel wrote:
What is the area of the triangle ABC?
(A) 15
(B) \(15\sqrt{3}\)
(C) \(\frac{5\sqrt{119}}{2}\)
(D) 32.5
(E) 36
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Draw a perpendicular from point C to meet side AB at point D
CDB is a 30-60-90 triangle with
∠C = 30 deg, ∠D = 90 deg & ∠B = 60 deg
--> BD: CD: BC = 1: \(\sqrt{3}\): 2
--> BD = k, CD = \(\sqrt{3}\)k, BC = 2k
Given BC = 2k = 5
--> k = 5/2
Altitude = \(\sqrt{3}\)k = 5/2*\(\sqrt{3}\)
Area of triangle ABC = 1/2*base*height = 1/2*AB*CD
= 1/2*12*5/2*\(\sqrt{3}\)
= 15\(\sqrt{3}\)
IMO Option B
Pls Hit Kudos if you like the solution
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66% (02:00) correct 
