In a right-angled triangle ABC, the angle at B is 90°. If AB = 6 cm and BC = 8 cm, what is the distance between the centroid G and the orthocenter H?
\(\frac{10}{3}\) cm
In right-angled triangle ABC the right angle is at B, with AB = 6 cm and BC = 8 cm. We need the distance between the centroid G and the orthocentre H.
In any right triangle the orthocentre (meeting point of the altitudes) lies at the right-angle vertex, so H coincides with B.
The circumcentre O lies at the midpoint of the hypotenuse. The hypotenuse \(AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{6^{2}+8^{2}}=\sqrt{36+64}=\sqrt{100}=10\) cm.
The distance from B (the right angle) to the midpoint of AC equals the circumradius, so \(OH=\frac{AC}{2}=\frac{10}{2}=5\) cm.
On the Euler line the centroid G divides the segment from orthocentre H to circumcentre O in the ratio 2 : 1, so \(HG=\frac{2}{3}HO\).
Therefore \(HG=\frac{2}{3}\times 5=\frac{10}{3}\) cm.
The key facts are the position of the orthocentre and circumcentre in a right triangle and the Euler-line 2:1 division. Hence the distance between the centroid and the orthocentre is 10/3 cm.
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