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Question Number 200102 by Blackpanther last updated on 13/Nov/23

Answered by som(math1967) last updated on 14/Nov/23

$$EismidptofACandAB\parallel DE$$$$\therefore DismidptofBC$$$$2ar△ABD=ar△ABC$$$$AB\parallel DE\therefore ar△ADE=ar△ABD$$$$ar△ADE=△AEF+△EFD$$$$=12+6=18squnit$$$$ar△ABC=2\times 18=36squnit$$

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