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.

Guest Sep 29, 2020

#1**+1 **

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!**

\(A=\frac{1}{2}ah\)

\(h=\sqrt{3}\)

\(a^2=(\frac{a}{2})^2+h^2\\ a^2-\frac{a^2}{4}=h^2\)

\(\frac{3}{4}a^2=3\\ a=\sqrt{\frac{3\cdot 4}{3}}\)

\(a=2\)

\(A=\frac{1}{2}ah\)

\(A=\frac{1}{2}\cdot 2\cdot \sqrt{3}\)

\(A=\sqrt{3}\)

!

asinus Sep 29, 2020

#2**0 **

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

ElectricPavlov Sep 29, 2020

#3**+1 **

SORRY .... I see your answer method now ! I read the question incorrectly as an isosceles triangle rather than __ equilateral __triangle .... D'Oh !

ElectricPavlov
Sep 29, 2020