SSC CHSL 23 March 2018 Evening Shift Question 5

Question 5

A line DE parallel to the side BC intersects the other two sides of triangle at point D and E such that AD=$$\frac{1}{6}$$ AB and AE=$$\frac{1}{6}$$ AC. If the value of BC is 18 cm then calculate the value of DE (in cm).


Given : AD=$$\frac{1}{6}$$ AB and AE=$$\frac{1}{6}$$ AC and BC = 18 cm

To find : DE = ?

Solution : In $$\triangle$$ ADE and $$\triangle$$ ABC,

$$\angle$$ A = $$\angle$$ A     (Common Angle)

$$\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{6}$$     (Given)

$$\therefore$$ $$\triangle$$ ADE $$\sim$$ $$\triangle$$ ABC

=> $$\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$$

=> $$\frac{DE}{18}=\frac{1}{6}$$

=> $$DE=\frac{18}{6}=3$$ cm

=> Ans - (B)

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