D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.

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SSC CGL 2023 Tier-I Official Paper (Held On: 18 Jul 2023 Shift 3)
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  1. 49 ∶ 144
  2. 49 ∶ 81
  3. 49 ∶ 256  
  4. 256 ∶ 49

Answer (Detailed Solution Below)

Option 3 : 49 ∶ 256  
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Detailed Solution

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Given:

AD : DB = 7 : 9

DE ∥ BC

Calculation:

F1 SSC Arbaz 6-10-23 D12

AD : DB = 7 : 9

So, AD : AB = 7 : (7 + 9) = 7 : 16

In ΔADE and ΔABC

∠ ADE = ∠ ABC    --- (corresponding angle)

∠ AED = ∠ ACB   --- (corresponding angle)

∠ A = ∠ A       --- (common angle)

So,  ΔADE ∼ ΔABC

AD : AB = DE : BC = 7 : 16

Now,

In ΔDEF and ΔBCF

∠ DEF = ∠ FBC    ---(alternate angle)

∠ EDF = ∠ FCB    ---(alternate angle)

∠ DFE = ∠ BFC   --- (Opposite Angles)

So, In ΔDEF ∼  ΔBCF

Thus, areas of ΔDEF and ΔCBF

⇒ DE2 : BC2

72 : 162

49 : 256

∴ The correct option is 3

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