Question 62

$$\triangle$$ABC $$\sim$$ $$\triangle$$EDF and ar($$\triangle$$ABC) : ar($$\triangle$$DEF) = 4 : 9. If AB = 6 cm, BC = 8 cm and AC = 10 cm,then DF is equal to:

Solution

We have
ABC similar to EDF
Now ratio of areas of similar triangles = (Square of ratio of sides )
Therefore we can say
The ratio of sides of ABC and EDF will be $$\sqrt{\ \frac{4}{9}\ }=\ \frac{2}{3}$$
Now we get
BC : DF = 2:3
Substituting we get DF = 12

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