1) It is an equilateral triangle. The Pythagorean Theorem will not do you much direct good.
2) If you construct a perpendicular on each of the three sides, through point P, you should start seeing Right Triangles in your dreams.
Let ABC be an equilateral triangle and P be a point inside this triangle such that PA=x, PB=y and PC=z.
If , find the length of the sides of triangle ABC in terms of x and y.
How do we go about this? AFAIK, the Pythagorean theorem can be applied to Right Angle Triangles. I am sure there is a Right Angle Triangle hidden somewhere in there...
1) It is an equilateral triangle. The Pythagorean Theorem will not do you much direct good.
2) If you construct a perpendicular on each of the three sides, through point P, you should start seeing Right Triangles in your dreams.
Say you take those 3 : PA, PB and PC, and form a triangle with them, what will you get?
A right triangle, agree? If not, then all I can say is you need basic help from your teacher.
OK, do some work:
draw equilateral triangle ABC, insert point P inside so that AP=3, CP=4 and BP = 5; like:
x^2 + y^2 = z^2Code:A x=3 P y=4 z=5 C B
3^2 + 4^2 = 5^2
9 + 16 = 25
25 = 25 ; got that?
Now go calculate the side length of triangle ABC.
Once you "see" what's going on, do over in terms of x,y,z.
Well, before you edited ur post, I actually did some research on it.
As a general rule, for every equilateral triangle of side , the area is given by
If we make a point P at any random location in the triangle and draw three lines through it each perpendicular to a side, we end up in something like this (ignore the side lengths 1. Just assume them to be )
Now the area of each separate triangle formed is
x a x h1
x a x h2 and
x a x h3
respectively.
Hence the equation:
= x a x h1 + x a x h2 + x a x h3
Multiplying both sides by :
= h1 + h2 + h3
in our case...
= x + y + z
and the second equation is:
Now we can solve the two simultaneously to obtain a.
BUT
there are infinite values which satisfy the equation
Putting those values in the equation = x + y + z yields infinite values for a also.