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Question-200102




Question Number 200102 by Blackpanther last updated on 13/Nov/23
Answered by som(math1967) last updated on 14/Nov/23
 E is mid pt of AC and AB∥DE  ∴ D is mid pt of BC  2ar△ABD=ar△ABC   AB∥DE ∴ ar△ADE=ar△ABD  ar △ADE=△AEF+△EFD    =12+6=18squnit  ar △ABC=2×18=36sq unit
$$\:{E}\:{is}\:{mid}\:{pt}\:{of}\:{AC}\:{and}\:{AB}\parallel{DE} \\ $$$$\therefore\:{D}\:{is}\:{mid}\:{pt}\:{of}\:{BC} \\ $$$$\mathrm{2}{ar}\bigtriangleup{ABD}={ar}\bigtriangleup{ABC} \\ $$$$\:{AB}\parallel{DE}\:\therefore\:{ar}\bigtriangleup{ADE}={ar}\bigtriangleup{ABD} \\ $$$${ar}\:\bigtriangleup{ADE}=\bigtriangleup{AEF}+\bigtriangleup{EFD} \\ $$$$\:\:=\mathrm{12}+\mathrm{6}=\mathrm{18}{squnit} \\ $$$${ar}\:\bigtriangleup{ABC}=\mathrm{2}×\mathrm{18}=\mathrm{36}{sq}\:{unit} \\ $$

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