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Question-181329




Question Number 181329 by manxsol last updated on 24/Nov/22
Answered by som(math1967) last updated on 24/Nov/22
Commented by som(math1967) last updated on 24/Nov/22
△AOD∼△COB   ((AO)/(CO))=((DO)/(BO))=((AD)/(BC))=((16)/4)=(4/1)  AC=10⇒AO=(4/5)×10=8  ((AO+DO+AD)/2)=((8+11.2+16)/2)=17.6  DO=14⇒DO=(4/5)×14=11.2  Ar of△AOD=(√(17.6×1.6×6.4×9.6))                                  =41.6 (approx)  ((△BOA)/(△AOD))=((BO)/(DO))=(1/4)  ⇒△BOA=((41.6)/4)=10.4=△COD   ((△BOC)/(△BOA))=(1/4)⇒△BOC=((10.4)/4)=2.6  Area of ADCB=41.6+2×10.4+2.6  =65sq unit
$$\bigtriangleup{AOD}\sim\bigtriangleup{COB} \\ $$$$\:\frac{{AO}}{{CO}}=\frac{{DO}}{{BO}}=\frac{{AD}}{{BC}}=\frac{\mathrm{16}}{\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{1}} \\ $$$${AC}=\mathrm{10}\Rightarrow{AO}=\frac{\mathrm{4}}{\mathrm{5}}×\mathrm{10}=\mathrm{8} \\ $$$$\frac{{AO}+{DO}+{AD}}{\mathrm{2}}=\frac{\mathrm{8}+\mathrm{11}.\mathrm{2}+\mathrm{16}}{\mathrm{2}}=\mathrm{17}.\mathrm{6} \\ $$$${DO}=\mathrm{14}\Rightarrow{DO}=\frac{\mathrm{4}}{\mathrm{5}}×\mathrm{14}=\mathrm{11}.\mathrm{2} \\ $$$${Ar}\:{of}\bigtriangleup{AOD}=\sqrt{\mathrm{17}.\mathrm{6}×\mathrm{1}.\mathrm{6}×\mathrm{6}.\mathrm{4}×\mathrm{9}.\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{41}.\mathrm{6}\:\left({approx}\right) \\ $$$$\frac{\bigtriangleup{BOA}}{\bigtriangleup{AOD}}=\frac{{BO}}{{DO}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\bigtriangleup{BOA}=\frac{\mathrm{41}.\mathrm{6}}{\mathrm{4}}=\mathrm{10}.\mathrm{4}=\bigtriangleup{COD} \\ $$$$\:\frac{\bigtriangleup{BOC}}{\bigtriangleup{BOA}}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\bigtriangleup{BOC}=\frac{\mathrm{10}.\mathrm{4}}{\mathrm{4}}=\mathrm{2}.\mathrm{6} \\ $$$${Area}\:{of}\:{ADCB}=\mathrm{41}.\mathrm{6}+\mathrm{2}×\mathrm{10}.\mathrm{4}+\mathrm{2}.\mathrm{6} \\ $$$$=\mathrm{65}{sq}\:{unit} \\ $$$$ \\ $$
Commented by manxsol last updated on 24/Nov/22
Yes,Sr.Som  atrap(B,b,d1,d2) is equal  heron(B+b,d1,d2)=64.99≈65
$${Yes},{Sr}.{Som} \\ $$$${atrap}\left({B},{b},{d}\mathrm{1},{d}\mathrm{2}\right)\:{is}\:{equal} \\ $$$${heron}\left({B}+{b},{d}\mathrm{1},{d}\mathrm{2}\right)=\mathrm{64}.\mathrm{99}\approx\mathrm{65} \\ $$

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