The sides of a triangle are k, 1·5k and 2·25k. What is the sum of the squares of its medians?

Given:

The sides of a triangle are k, 1.5k, and 2.25k.

Formula used:

The sum of the squares of medians of a triangle is given by:

\(\text{Sum of squares of medians} = \frac{3}{4} (a^2 + b^2 + c^2)\)

Where, a, b, and c are the sides of the triangle.

Calculation:

Let a = k, b = 1.5k = 3k/2, c = 2.25k = 9k/4.

\(a^2 + b^2 + c^2 = k^2 + \frac{9}{4}k^2 + \frac{81}{16}k^2\)

\(a^2 + b^2 + c^2 = \frac{16}{16}k^2 + \frac{36}{16}k^2 + \frac{81}{16}k^2 = \frac{133}{16}k^2\)

Sum of squares of medians = \(\frac{3}{4} \times \frac{133}{16}k^2\)

Sum of squares of medians = \(\frac{3 \times 133}{4 \times 16}k^2 = \frac{399}{64}k^2\)

Therefore, the sum of the squares of the medians is: \(\frac{399}{64}k^2\)