For a right angle triangle the following equation is true:
a^2 + b^2 = c^2
Where c is the hypotenuse of the triangle and a,b are the other sides (you decide which side is a and which side is b).
I am trying to solve the length of the hypotenuse of a right triangle with sides of 6mm and 8mm
4 answers
This involves the pythagorean theorem. Let c = hypotenuse; a = 1 side; b = other side.
c^2 = a^2 + b^2
c^2 = 6^2 + 8^2
c^2 = 36 + 64
Solve for c.
c^2 = a^2 + b^2
c^2 = 6^2 + 8^2
c^2 = 36 + 64
Solve for c.
I am trying to solve the length of the hypotenuse of a right triangle with sides of 6mm and 8mm
a = 6mm, b = 8mm
For b > a
c = b^2[1 + a^2/b^2]/sqrt(a^2 + b^2)
a = 6mm, b = 8mm
For b > a
c = b^2[1 + a^2/b^2]/sqrt(a^2 + b^2)
Regarding:
For b > a
c = b^2[1 + a^2/b^2]/sqrt(a^2 + b^2)
I forgot to acknowledge that when this expression is multipied out, the end result is c = sqrt(a^2 + b^2).
I was initially surprised when deriving it from the relationship between the altitude to the hypotenuse and the hypotenuse itself;
* The altitude to the hypotenuse of a right triangle is the geometric mean between the segments of the hypotenuse created by the point where the altitude intersects the hypotenuse or h^2 = xy.
* With an altitude drawn to the hypotenuse of a right triangle, each leg is the geometric mean between the hypotenuse and the segment of the hypotenuse in contact with the leg.
Can't beat old Pythagorus.
For b > a
c = b^2[1 + a^2/b^2]/sqrt(a^2 + b^2)
I forgot to acknowledge that when this expression is multipied out, the end result is c = sqrt(a^2 + b^2).
I was initially surprised when deriving it from the relationship between the altitude to the hypotenuse and the hypotenuse itself;
* The altitude to the hypotenuse of a right triangle is the geometric mean between the segments of the hypotenuse created by the point where the altitude intersects the hypotenuse or h^2 = xy.
* With an altitude drawn to the hypotenuse of a right triangle, each leg is the geometric mean between the hypotenuse and the segment of the hypotenuse in contact with the leg.
Can't beat old Pythagorus.