Question 168

ΔABC is similar to ΔDEF. The ratio of their perimeters is 4 :1. The ratio of their areas is

Solution

$$\triangle$$ABC $$\sim$$ $$\triangle$$DEF

$$\therefore$$ $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$

=> $$\frac{AB+BC+AC}{DE+EF+DF} = \frac{4}{1}$$

$$\therefore$$ Ratio of area of $$\triangle$$ABC to that of $$\triangle$$DEF

= $$\frac{AB^2}{DE^2} = \frac{16}{1}$$

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