Problem: If the altitude of an equilateral triangle is 6\sqrt{6}6, then the area is:
Answer Choices:
A. 222 \sqrt{2}22
B. 232 \sqrt{3}23
C. 333 \sqrt{3}33
D. 626 \sqrt{2}62
E. 121212 Solution:
h=s23;∴s=2h3,A=s234=4h23⋅34=633=23h=\dfrac{s}{2} \sqrt{3} ; \quad \therefore s=\dfrac{2 h}{\sqrt{3}}, A=\dfrac{s^{2} \sqrt{3}}{4}=\dfrac{4 h^{2}}{3} \cdot \dfrac{\sqrt{3}}{4}=\dfrac{6 \sqrt{3}}{3}=2 \sqrt{3}h=2s3;∴s=32h,A=4s23=34h2⋅43=363=23.