a right angled triangle has legs a,b (a is greater than b).the right angle is bisected splitting the original triangle into 2 smaller triangles.Find the dist btw. the orthocentres of the smaller triangles.
Anonymous/Sania/Sonia -- please use the same name for your posts.
no offense but it is kind of hard to solve a problem to find a length when given no lengths...did you copy/paste wrong? Anyway, first when there is an angle bisection always use angle bisector theorem. also, the orthocenter is where the altitudes meet
To find the distance between the orthocentres of the two smaller triangles formed by bisecting the right angle of a right-angled triangle, we need to understand the properties of orthocentres and how the triangles are divided.
First, let's define the orthocentre of a triangle. The orthocentre is the point where the three altitudes of a triangle intersect. An altitude is a line segment drawn from a vertex of a triangle perpendicular to the opposite side.
In the original right-angled triangle with legs a and b (where a is greater than b), suppose the legs meet at a right angle at vertex C. Let the length of the hypotenuse be c.
Now, bisect the right angle at C to form two smaller triangles. Let's call the two new vertices where the bisecting line intersects the legs as D and E. Triangle ACD and triangle BCE are the two smaller triangles.
To find the orthocentre of each smaller triangle, we need to first find the altitudes.
In triangle ACD, the altitude from vertex D is perpendicular to AC. Similarly, in triangle BCE, the altitude from vertex E is perpendicular to BC.
Since the original triangle is a right-angled triangle, the altitude from D and E will be the same as the legs.
The altitude from vertex D is the segment DE, which has a length of b.
The altitude from vertex E is the segment DE, which also has a length of b.
Therefore, the orthocentres of the smaller triangles will coincide at the midpoint of segment DE.
To find the distance between the orthocentres, we need to calculate the length of DE.
Since DE is the perpendicular bisector of the hypotenuse AC, it divides AC into two congruent segments.
Using the properties of right-angled triangles, we know that AC is the hypotenuse, and the length of the hypotenuse can be found using the Pythagorean theorem:
c^2 = a^2 + b^2
Now, since DE is the perpendicular bisector of AC, it divides AC into two segments of equal length. Therefore, the length of each segment will be c/2.
Finally, to find the distance between the orthocentres, it is equal to the length of segment DE, which is c/2.
In conclusion, the distance between the orthocentres of the two smaller triangles formed by bisecting the right angle of a right-angled triangle is half the length of the hypotenuse of the original triangle.