Question 62

$$\triangle$$ABC $$\sim$$ $$\triangle$$EDF and ar($$\triangle$$ABC): ar($$\triangle$$DEF) = 1 : 4. If AB = 7cm, BC = 8 cm and CA = 9 cm, then DF is equal to:

Solution

Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides. This proves that the ratio of areas of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.

Since $$\triangle$$ABC $$\sim$$ $$\triangle$$EDF ,

$$\frac{area of triangle$$ABC}{area of triange EDF} = \frac{BC^2}{DF^2}$$

$$\frac{1}{2} = \frac{8}{DF}$$

DF = 16 cm

So , the answer would be option b)16 cm.


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