Question

If the orthocenter, centroid, incenter and circumcentre coincide in a triangle $ABC$, and if the length of side $AB$ is $\sqrt{75}$ units, then the length of the altitude of the triangle through the vertex $A$ is:

• A. $\sqrt{3}$ units
• B. $3$ units
• C. $\frac{\sqrt{15}}{2}$ units
• D. $\frac{15}{2}$ units
• E. $\frac{\sqrt{5}}{2}$ units

Hint:

In any equilateral triangle with side $a$, the altitude, median, and perpendicular bisector are the same segment with length $a\sin(60^\circ)$. Always simplify square roots like $\sqrt{75}$ early to make the multiplication cleaner.

Answer 1

The Correct Option is D

Concept: A triangle where the orthocenter, centroid, incenter, and circumcenter all coincide is necessarily an equilateral triangle. In an equilateral triangle, all sides are equal, and the altitude (which also acts as the median and angle bisector) can be calculated using the side length.
• Properties: All angles are $60^\circ$.
• Altitude formula: $h = \frac{\sqrt{3}}{2} \times \text{side}$.

Step 1: Identify the type of triangle and side length.
Because all center points coincide, triangle $ABC$ is equilateral. The side length $s$ is given as $AB = \sqrt{75}$ units. We can simplify $\sqrt{75}$: \[ s = \sqrt{25 \times 3} = 5\sqrt{3} \text{ units} \]

Step 2: Apply the altitude formula.
The altitude $h$ through vertex $A$ is: \[ h = \frac{\sqrt{3}}{2} \times s \] Substituting $s = 5\sqrt{3}$: \[ h = \frac{\sqrt{3}}{2} \times 5\sqrt{3} \]

Step 3: Calculate the final value.
\[ h = \frac{5 \times (\sqrt{3} \times \sqrt{3})}{2} = \frac{5 \times 3}{2} = \frac{15}{2} \text{ units} \]