Question Number 221870 by fantastic last updated on 11/Jun/25
Answered by mehdee7396 last updated on 12/Jun/25
$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{AB}×{OM}\:\:\:\:\&\:\:\:{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{CD}×{ON} \\ $$$${AB}={CD}\:\:\Rightarrow{S}_{\mathrm{1}} +{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{AB}×\left({OM}+{ON}\right)=\frac{{AB}×{MN}}{\mathrm{2}}=\mathrm{16} \\ $$$$\Rightarrow{S}={AB}×{MN}=\mathrm{32} \\ $$
Commented by mehdee7396 last updated on 12/Jun/25
Answered by fantastic last updated on 12/Jun/25
Commented by fantastic last updated on 12/Jun/25
$${PQ}\:\parallel{AB}\parallel{CD}.{PQ}\:{is}\:{drawn}\:{from}\:{O} \\ $$$${Both}\:\bigtriangleup{COD}\:{and}\:{parallelogram}\:{PDCQ}\:{have}\:{the}\:{same}\:{base}\:{CD}\: \\ $$$${and}\:{point}\:{O}\:{is}\:{touching}\:{PQ} \\ $$$${So}\:\bigtriangleup{COD}=\frac{\mathrm{1}}{\mathrm{2}}\Box{PDCQ} \\ $$$${Similarly}\:\bigtriangleup{AOB}=\frac{\mathrm{1}}{\mathrm{2}}\Box{APQB} \\ $$$${So}\:\Box{ABCD}=\Box{PDCQ}+\Box{APQB} \\ $$$$=\mathrm{2}\left(\bigtriangleup{COD}+\bigtriangleup{AOB}\right) \\ $$$$=\mathrm{2}×\mathrm{16} \\ $$$$=\mathrm{32}{sq}.{cm} \\ $$$$\: \\ $$