 Posers and Puzzles wolfgang59 Posers and Puzzles 14 Oct '07 14:57
1. 14 Oct '07 14:57
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. 🙂
2. 14 Oct '07 16:38
Originally posted by wolfgang59
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. 🙂
Divide the triangle in two equal parts.

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?
3. 14 Oct '07 17:36
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!!
4. 14 Oct '07 19:55
Originally posted by wolfgang59
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!!
Ah, that is also a nice solution 🙂
5. 16 Oct '07 15:062 edits
Originally posted by wolfgang59
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!!
Really elegant. Also...it can be viewed as being equal to the area of an equilateral triangle of side 12/(sqrt(3)).
The two halves of this equilateral triangle having been juxtaposed to form the given isosceles triangle with vertex angle 120 deg...