Question 41
In $$\triangle ABC, AB = 6 cm, AC = 8 cm,$$ and $$BC = 9 cm$$. The length of median $$AD$$ is:
Solution
In ∆ABC, AD is median . D. will be the mid point of BC.
BD = CD = BC/2= 9/2 = 4.5 cm
$$AB^2 + AC^2 = 2(AD^2 + BD^2)$$
$$6^2 + 8^2 = 2(AD^2 + (4.5)^2)
$$100/2 = AD^2 + 20.25$$
$$AD^2 = 29.75 = 119/4$$
$$AD = \frac{\sqrt{119}}{2}$$
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