Q. 1

# , , are the medians of ΔABC. If BE = 12, CF = 9 and AB^{2} + BC^{2} + AC^{2} = 600, BC = 10, find AD.

Answer :

From the given info,

As AB^{2} + BC^{2} + AC^{2} = 600

and BC = 10,

⇒ AB^{2} + AC^{2} = 500 ...... (1)

As AD is median,

⇒ BD = 5 ...... (2)

In Δ ABC, using the theorem of Apolloneous, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

Substituting values for (1) and (2) in above equation,

⇒ 500 = 2 (AD^{2} + 5^{2})

⇒ 250 = AD^{2} + 25

⇒ AD^{2} = 225

⇒ AD^{2} = 15^{2}

⇒ AD = 15

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