Two vertices of an equilateral triangle are origin and (4, 0). What is the area of the triangle? |
A) \[4\] sq units
B) \[\sqrt{3}\]sq units
C) \[4\sqrt{3}\] sq units
D) \[2\sqrt{3}\] sq units
Correct Answer: C
Solution :
Since, triangle is equilateral. |
\[\therefore \]\[AB=BC=CA\]\[\Rightarrow \]\[BD=\frac{4}{2}=2\] |
In \[\Delta ADC,\]\[A{{D}^{2}}={{4}^{2}}-{{2}^{2}}=16-4=12\] |
\[\Rightarrow \] \[AD=2\sqrt{3}\] |
\[\therefore \]Area of \[\Delta ABC=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times 4\times 2\sqrt{3}\] |
\[=4\sqrt{3}\,sq\text{ }units\] |
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