Question

problem 1
consider this diagram.
what is the area of triangle abc?
a ( 6sqrt{3} ) square units
b ( 18sqrt{3} ) square units
c ( 36sqrt{3} ) square units
d ( 72sqrt{3} ) square units

Explanation:

Step1: Find the length of \(BD\)

In right - triangle \(BCD\), \(\cos60^{\circ}=\frac{BD}{BC}\). Given \(BC = 12\), and \(\cos60^{\circ}=\frac{1}{2}\).
So, \(BD=BC\times\cos60^{\circ}=12\times\frac{1}{2} = 6\).

Step2: Find the length of \(CD\)

In right - triangle \(BCD\), \(\sin60^{\circ}=\frac{CD}{BC}\). Given \(BC = 12\), and \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\).
So, \(CD=BC\times\sin60^{\circ}=12\times\frac{\sqrt{3}}{2}=6\sqrt{3}\).

Step3: Calculate the area of \(\triangle ABC\)

Since \(\triangle ABC\) is isosceles (\(\angle A=\angle B = 60^{\circ}\), \(AD = BD\)), \(AB=2BD = 12\).
The area formula for a triangle is \(A=\frac{1}{2}\times base\times height\). Here, base \(AB = 12\) and height \(CD=6\sqrt{3}\).
\(A=\frac{1}{2}\times AB\times CD=\frac{1}{2}\times12\times6\sqrt{3}=36\sqrt{3}\).

Answer:

C. \(36\sqrt{3}\) square units