If the sides of a right triangle are $9$ cm, $12$ cm, and $15$ cm long, then the sum of squares of medians is:

227.5

^{2}

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337.5

^{2}

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537.5

^{2}

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None of these

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Solution

The correct option is D $337.5$ cm $^{2}$
We know the formulae,

$M e d i a n, (m_{1})^{2} = \frac{1}{4} (2 (s i d e_{2})^{2} + 2 (s i d e_{3})^{2} - (s i d e_{1})^{2})$

$M e d i a n, (m_{2})^{2} = \frac{1}{4} (2 (s i d e_{3})^{2} + 2 (s i d e_{1})^{2} - (s i d e_{2})^{2})$

$M e d i a n, (m_{3})^{2} = \frac{1}{4} (2 (s i d e_{1})^{2} + 2 (s i d e_{2})^{2} - (s i d e_{3})^{2})$

Let, $s i d e_{1} = 9, s i d e_{2} = 12, s i d e_{3} = 15$

$(m_{1})^{2} = \frac{1}{4} (2 (12)^{2} + 2 (15)^{2} - 9^{2}) = \frac{1}{4} (288 + 450 - 81) = \frac{657}{4}$

$(m_{2})^{2} = \frac{1}{4} (2 (15)^{2} + 2 (9)^{2} - 12^{2}) = \frac{1}{4} (450 + 162 - 144) = \frac{468}{4}$

$(m_{3})^{2} = \frac{1}{4} (2 (9)^{2} + 2 (12)^{2} - 15^{2}) = \frac{1}{4} (162 + 288 - 225) = \frac{225}{4}$

$(m_{1})^{2} + (m_{2})^{2} + (m_{3})^{2} = \frac{657}{4} + \frac{468}{4} + \frac{225}{4} = \frac{657 + 468 + 225}{4} = \frac{1350}{4}$

$(m_{1})^{2} + (m_{2})^{2} + (m_{3})^{2} = 337.5 s q . c m$

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