Question

What is the area of an equilateral triangle with a side of 8?

Answer 1

The area of a equilateral triangle with sides a is

`A=sqrt3/4*a^2=>A=sqrt3/4*(8)^2=27.71`

Answer 2

Area equals to `16sqrt(3)`

Explanation:

Consider an equilateral triangle `Delta ABC`:

The area of this triangle is
`S=1/2*b*h`

All its sides are given and equal to `8`:
`a=b=c=8`,
its altitude `h` is not given, but can be calculated

Let the base of the altitude from vertex `B` to side `AC` be point `P`. Consider two right triangles `Delta ABP` and `Delta CBP`. They are congruent by a common cathetus `BP` and congruent hypotenuses `AB=c=BC=a`.
Therefore, the other pair of catheti, `AP` and `CP` are congruent as well:
`AP=CP=b/2`

Now the altitude `BP=h` can be calculated from the Pythagorean Theorem applied to a right triangle `Delta ABP`:
`c^2 = h^2 + (b/2)^2`
from which
`h=sqrt(c^2-(b/2)^2)=sqrt(64-16)=4sqrt(3)`

Now the area of triangle `Delta ABC` can be determined:
`S=1/2*8*4sqrt(3)=16sqrt(3)`

Answer 3

16`sqrt`3

Explanation:

Area of equilateral triangle = `sqrt3 a^2`/4
In this situation,
Area = `sqrt3*8^2`/4
= `sqrt3*64`/4
= `sqrt3*16`
= 16`sqrt3` sq. unit