14 Oct 07
Originally posted by wolfgang59Divide the triangle in two equal parts.
Here is an easy geometry puzzle for those that cannot do Ranjan's stinkers!
What is the area of an isoceles triangle with a base of 12 and an apex of 120 degrees.
Elegant and short proof required. 🙂
Each of the halves is a 30-60-90 triangle with middle side 6.
This means that the short side (wich is the height of the original triangle) of the halves is 6/sqrt(3).
So the area of the original triangle is 1/2 * 12 * 6/sqrt(3) = 12 * sqrt(3)
Elegant enough?
14 Oct 07
Correct but you have assumed the reader knows ratios of the sides of a 30, 60, 90 triangle.
I thought a neater idea was to spot that the described triangle is a third of an equilateral triangle of side 12.
So using your shortcut we have area of equilateral triangle is
0.5 * 12 * 6sqrt3 = 36sqrt3
Therefore isoceles is 12sqrt3
I said it was easy!!
14 Oct 07
Originally posted by wolfgang59Ah, that is also a nice solution 🙂
Correct but you have assumed the reader knows ratios of the sides of a 30, 60, 90 triangle.
I thought a neater idea was to spot that the described triangle is a third of an equilateral triangle of side 12.
So using your shortcut we have area of equilateral triangle is
0.5 * 12 * 6sqrt3 = 36sqrt3
Therefore isoceles is 12sqrt3
I said it was easy!!
16 Oct 07
2 edits
Originally posted by wolfgang59Really elegant. Also...it can be viewed as being equal to the area of an equilateral triangle of side 12/(sqrt(3)).
Correct but you have assumed the reader knows ratios of the sides of a 30, 60, 90 triangle.
I thought a neater idea was to spot that the described triangle is a third of an equilateral triangle of side 12.
So using your shortcut we have area of equilateral triangle is
0.5 * 12 * 6sqrt3 = 36sqrt3
Therefore isoceles is 12sqrt3
I said it was easy!!
The two halves of this equilateral triangle having been juxtaposed to form the given isosceles triangle with vertex angle 120 deg...