Question 76

$$\triangle ABC$$ is similar to $$\triangle DEF$$. If area of $$\triangle ABC$$ is 9 sq.cm. and area of $$\triangle DEF$$ is 16 sq.cm. and BC = 2.1 cm. Then the length of EF will be

Solution

if triangle are similar then 

$$\frac{area of \triangle ABC}{area of\triangle DEF}$$ = $$\frac{BC^2}{EF^2}$$ 

$$\frac{9}{16}$$ = $$\frac{BC^2}{EF^2}$$ 

$$\frac{2.1}{EF}$$  = $$\frac{3}{4}$$

EF = 2.8 cm


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