Question

In a triangle BC, if the mid points of sides AB, BC, CA are (3,0,0), (0,4,0),(0,0,5) respectively, then AB² + BC² + CA² =

• A. 50
• B. 200
• C. 300
• D. 400

Answer 1

The Correct Option is D

To solve the problem, we need to find the sum of the squares of the side lengths of triangle ABC given the midpoints of its sides.

1. Define the Vertices and Midpoints:
Let the vertices of triangle ABC be A = (x₁, y₁, z₁), B = (x₂, y₂, z₂), C = (x₃, y₃, z₃).
Let D, E, F be the midpoints of BC, CA, AB respectively, given as D = (3, 0, 0), E = (0, 4, 0), F = (0, 0, 5).

2. Set Up Midpoint Equations:
For D = ((x₂+x₃)/2, (y₂+y₃)/2, (z₂+z₃)/2) = (3, 0, 0):
x₂ + x₃ = 6, y₂ + y₃ = 0, z₂ + z₃ = 0.
For E = ((x₃+x₁)/2, (y₃+y₁)/2, (z₃+z₁)/2) = (0, 4, 0):
x₃ + x₁ = 0, y₃ + y₁ = 8, z₃ + z₁ = 0.
For F = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) = (0, 0, 5):
x₁ + x₂ = 0, y₁ + y₂ = 0, z₁ + z₂ = 10.

3. Solve for x-Coordinates:
From x₁ + x₂ = 0, x₁ = -x₂.
From x₃ + x₁ = 0, x₃ = -x₁ = x₂.
From x₂ + x₃ = 6, x₂ + x₂ = 6, so 2x₂ = 6, x₂ = 3.
Thus, x₁ = -3, x₃ = 3.

4. Solve for y-Coordinates:
From y₁ + y₂ = 0, y₁ = -y₂.
From y₃ + y₁ = 8, y₃ = 8 - y₁ = 8 + y₂.
From y₂ + y₃ = 0, y₂ + 8 + y₂ = 0, so 2y₂ = -8, y₂ = -4.
Thus, y₁ = 4, y₃ = 4.

5. Solve for z-Coordinates:
From z₁ + z₂ = 10, z₁ = 10 - z₂.
From z₃ + z₁ = 0, z₃ = -z₁ = z₂ - 10.
From z₂ + z₃ = 0, z₂ + z₂ - 10 = 0, so 2z₂ = 10, z₂ = 5.
Thus, z₁ = 5, z₃ = -5.

6. Determine the Vertices:
The vertices are A = (-3, 4, 5), B = (3, -4, 5), C = (3, 4, -5).

7. Calculate the Side Lengths Squared:
AB² = (3 - (-3))² + (-4 - 4)² + (5 - 5)² = 6² + (-8)² = 36 + 64 = 100.
BC² = (3 - 3)² + (4 - (-4))² + (-5 - 5)² = 0 + 8² + (-10)² = 64 + 100 = 164.
CA² = (-3 - 3)² + (4 - 4)² + (5 - (-5))² = (-6)² + 0 + 10² = 36 + 100 = 136.

8. Sum the Squares:
AB² + BC² + CA² = 100 + 164 + 136 = 400.

Final Answer:
The sum of the squares of the side lengths is 400.