Triangle problem - solution
This is a solution to the triangle problem I posted here.
The picture is below; we are to find the area of the triangle.
Since AC is tangent to the circle at D and AB is tangent to the circle at E, then the distances AD and AE are equal. That is, AE must be 2.
I can't remember nor find a name for this theorem. It has to do with two tangents from the same point (and it is easy to prove using congruent triangles), saying that those distances are congruent.
Similarly, CD and CF are congruent -- both are 3. And also by the same reasoning, BE = BF (because BA and BC are tangets to the circle).
I will call BE and BF as x, it's easier to manipulate in an equation.
Now, we can solve x fairly easily by applying the Pythagorean theorem to the right triangle ABC.
(x + 2)2 + (x + 3)2 = (2 + 3)2
x2 + 4x + 4 + x2 + 6x + 9 = 25
2x2 + 10x + 13 = 25
2x2 + 10x − 12 = 0
x2 + 5x − 6 = 0
x = (−5 ± √( 25 − 4(−6))) / 2
x = (−5 ± 7) / 2
x = − 6 or x = 1. Obviously we choose x = 1.
Then, we go on to solve the area of the triangle. The two legs of the triangle are thus 3 and 4, and so the area is 3 × 4 / 2 = 6 square units.
All done!
The picture is below; we are to find the area of the triangle.
Since AC is tangent to the circle at D and AB is tangent to the circle at E, then the distances AD and AE are equal. That is, AE must be 2.
I can't remember nor find a name for this theorem. It has to do with two tangents from the same point (and it is easy to prove using congruent triangles), saying that those distances are congruent.
Similarly, CD and CF are congruent -- both are 3. And also by the same reasoning, BE = BF (because BA and BC are tangets to the circle).
I will call BE and BF as x, it's easier to manipulate in an equation.
Now, we can solve x fairly easily by applying the Pythagorean theorem to the right triangle ABC.
(x + 2)2 + (x + 3)2 = (2 + 3)2
x2 + 4x + 4 + x2 + 6x + 9 = 25
2x2 + 10x + 13 = 25
2x2 + 10x − 12 = 0
x2 + 5x − 6 = 0
x = (−5 ± √( 25 − 4(−6))) / 2
x = (−5 ± 7) / 2
x = − 6 or x = 1. Obviously we choose x = 1.
Then, we go on to solve the area of the triangle. The two legs of the triangle are thus 3 and 4, and so the area is 3 × 4 / 2 = 6 square units.
All done!
Comments
Since this right triangle has a hypotenuse of 5 the legs MUST BE 3 and 4.
So:
A=0.5bh
A=0.5*4*3
A=6 sq units
The solution is so elegant that I am led to think there must be a more elegant insight that would avoid the heavy algebra. If anyone finds such a solution I would like to see it.
Thank you for the information, really clever technique.
Dear Wally,
we can not predict the length of the right angle triangle be merely having the hypotenuse, we need another parameter (angle or another side)
best regards