Suzanna wrote:Can you prove that the area of an equilateral triangle is √3/4 * a2, where a is the side of the triangle?
There's a lot of ways. You can cut the triangle into two halves, rerrange it into a rectangle, and use Pythagoras.
You can also use the special properties of 30-60-90 triangles: Imagine an equilateral triangle sitting on its base whose sides are all "a". Draw a line from the apex (top vertex) down to the base. You have just cut the equilateral triangle into two identical right triangles whose hypotenuse is "a".
Let's look at one of these right triangles more closely. You can see that the base is a/2. Because this is a right triangle, and because the hypotenuse is exactly double the base, this must be a 30-60-90 triangle whose proportions are always:
x : x*root 3 : 2x
(short leg : long leg : hypotenuse).
Thus, the base of our right triangle is a/2 while the height is (a/2)*root3. Thus, using the common area formula for triangles ((base*height)/2), the area of our right triangle is: (1/2) * (a/2) *((a*root 3)/2) or ((a^2)*root 3)/8.
And, finally, because our original equilateral triangle was comprised of two of these right triangles, the area of the equilateral triangle must be double that.
Therefore, the area of an equilateral triangle is ((a^2) * root 3)/4 wherein "a" is a side of the triangle.
Note that papgust is 100% correct that you won't be expected to prove something like this. However, you will be expected to have facility with 30:60:90 triangles (which is why I went that route in the proof above).
Kaplan Teacher in Toronto
