Question

If the altitude of an equilateral triangle is $\sqrt{6}$, then its area is:

• A. $3\sqrt{3}$
• B. $2\sqrt{3}$
• C. $2\sqrt{2}$
• D. $6\sqrt{2}$

Answer 1

The correct option is B $2\sqrt{3}$

considering $2x$ as the side of the triangle

by drawing the altitude,we get a right triangle having sides $x$ and $\sqrt{6}$ and hypothenuse $2x$

by Phytagoras theorem,

$(2x{)}^2=x^2+(\sqrt{6}{)}^2$

$4x^2-x^2=6$

$3x^2=6$

$x^2=2$

$x=\sqrt{2}$

the side of the triangle is $2\sqrt{2}$

$A=\frac{2\sqrt{2}\times \sqrt{6}}{2}=\sqrt{12}=2\sqrt{3}$