The altitude of an equilateral triangle is the square root of 3 units. What is the area of the triangle, in square units? Express your answer in simplest radical form.
+14913 The altitude of an equilateral triangle is the square root of 3 units. What is the area of the triangle, in square units? Express your answer in simplest radical form.
Hello Guest!
\(A=\frac{1}{2}ah\)
\(h=\sqrt{3}\)
\(a^2=(\frac{a}{2})^2+h^2\\ a^2-\frac{a^2}{4}=h^2\)
\(\frac{3}{4}a^2=3\\ a=\sqrt{\frac{3\cdot 4}{3}}\)
\(a=2\)
\(A=\frac{1}{2}ah\)
\(A=\frac{1}{2}\cdot 2\cdot \sqrt{3}\)
\(A=\sqrt{3}\)
!
+36916 You lost me on this one, asinus.....
can you explain this step? :
a^2 = (a/2)^2 + h^2 won't that just give you the side length and not the base length?
Seems to me that we need more info to solve this Q 
+36916 SORRY .... I see your answer method now ! I read the question incorrectly as an isosceles triangle rather than equilateral triangle .... D'Oh !
