Question

In \(\triangle ABC\), mid-points of the sides AB, BC and CA are respectively \((l,0,0),(0,m,0)\) and \((0,0,n)\). Then \(\frac{(AB)^{2}+(BC)^{2}+(CA)^{2}}{l^{2}+m^{2}+n^{2}}=\)

• A. \(2\)
• B. \(16\)
• C. \(1\)
• D. \(8\)

Hint:

The segment joining two midpoints is half of the third side, so each side is twice a midpoint distance.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Step 2: Midpoint Distances
Midpoints of AB and BC: \((l,0,0)\) and \((0,m,0)\), distance \(\sqrt{l^{2}+m^{2}}=\dfrac{CA}{2}\).
Similarly, \(\dfrac{AB}{2}=\sqrt{m^{2}+n^{2}}\) and \(\dfrac{BC}{2}=\sqrt{l^{2}+n^{2}}\).

Step 3: Square the Sides
\(CA^{2}=4(l^{2}+m^{2})\), \(AB^{2}=4(m^{2}+n^{2})\), \(BC^{2}=4(l^{2}+n^{2})\).

Step 4: Add
Sum \(=8(l^{2}+m^{2}+n^{2})\), so the ratio is 8.

Final Answer:
The ratio equals 8. \[ \boxed{\text{(D) }8} \]