Question 67
In $$\triangle ABC, D, E$$ and $$F$$ are the midpoints of sides AB, BC and CA, respectively. If AB = 12 cm, BC = 20 cm and CA = 15 cm, then the value of $$\frac{1}{2}(DE + EF + DF)$$ is:
Solution
By mid point theorem,
DF = BC/2 = 20/2 = 10 cm
DE = AC/2 = 15/2 = 7.5 cm
EF = AB/2 = 12/2 = 6 cm
$$\dfrac{1}{2}(DE + EF + DF)$$
= $$\dfrac{1}{2}(7.5 + 6 + 10)$$
= $$\dfrac{23.5}{2}$$ = 11.75 cm
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