SSC CPO SI and ASI 2014 Paper

Question : 140 of 200

Marks: +1, -0

In a ∆ABC, D and E are two points on AB and AC respectively such that DE II BC, DE bisects the ∆ABC in two equal areas. Then the ratio DB : AB is

Solution:

(3) Given, DE ||BC

Area ΔABC = 2Area ΔADE and hence,

{\Δ{ADE}}/{\Δ{ABC}}= {1}/{2}= { {AD}^{2}}/{ {AB}^{2}}

So,

{ {AB}}/{ {AD}}=√{2} ⇒{ {AB}}/{ {AD}}-1

= { {AB}- {AD}}/{ {AD}}=√{2}-1

and

{D B}/{A D}= {D B}/{ {A B}/{√{2}}}=√{2}-1

⇒{D B}/{A B}= {√{2}-1}/{√{2}}

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