What is the base length of an equilateral triangle that has an area of approx 9.1 square centimeters?

1 Answer
Dec 1, 2015

~~4.584.58 cmcm

Explanation:

![http://jwilson.coe.uga.edu](https://useruploads.socratic.org/OVQ7ejRRq642nFyeGphv_kls1.jpg)

We can see that if we split an equilateral triangle in half, we are left with two congruent equilateral triangles. Thus, one of the legs of the triangle is 1/2s12s, and the hypotenuse is ss. We can use the Pythagorean Theorem or the properties of 30˚-60˚-90˚ triangles to determine that the height of the triangle is sqrt3/2s.

If we want to determine the area of the entire triangle, we know that A=1/2bh. We also know that the base is s and the height is sqrt3/2s, so we can plug those in to the area equation to see the following for an equilateral triangle:

A=1/2bh=>1/2(s)(sqrt3/2s)=(s^2sqrt3)/4

We know that the area of your equilateral triangle is 9.1.

We can set our area equation equal to 9.1:

9.1=(s^2sqrt3)/4

36.4=s^2sqrt3

s^2~~21.02

s~~4.58 cm