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Question-87803




Question Number 87803 by ajfour last updated on 06/Apr/20
Commented by ajfour last updated on 06/Apr/20
Nice question this is, posted by  Moth...(I haven′t tried but no  one else, either) mrW Sir shall  you, please? Area of green △=?
$${Nice}\:{question}\:{this}\:{is},\:{posted}\:{by} \\ $$$${Moth}…\left({I}\:{haven}'{t}\:{tried}\:{but}\:{no}\right. \\ $$$$\left.{one}\:{else},\:{either}\right)\:{mrW}\:{Sir}\:{shall} \\ $$$${you},\:{please}?\:{Area}\:{of}\:{green}\:\bigtriangleup=? \\ $$
Commented by john santu last updated on 06/Apr/20
9
$$\mathrm{9}\: \\ $$
Commented by ajfour last updated on 06/Apr/20
thanks, explain please.
$${thanks},\:{explain}\:{please}. \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
thank you for reposting the question sir
$${thank}\:{you}\:{for}\:{reposting}\:{the}\:{question}\:{sir} \\ $$
Answered by M±th+et£s last updated on 06/Apr/20
[ADM]+[MLB]=(1/2)[ABCD]=[AND]  ⇒([APM]+[APD])+(72+[QSTR]+8)  =[APD]+79+[QSTR]+10  [APM]=(79+10)−(72+8)  =9
$$\left[{ADM}\right]+\left[{MLB}\right]=\frac{\mathrm{1}}{\mathrm{2}}\left[{ABCD}\right]=\left[{AND}\right] \\ $$$$\Rightarrow\left(\left[{APM}\right]+\left[{APD}\right]\right)+\left(\mathrm{72}+\left[{QSTR}\right]+\mathrm{8}\right) \\ $$$$=\left[{APD}\right]+\mathrm{79}+\left[{QSTR}\right]+\mathrm{10} \\ $$$$\left[{APM}\right]=\left(\mathrm{79}+\mathrm{10}\right)−\left(\mathrm{72}+\mathrm{8}\right) \\ $$$$=\mathrm{9} \\ $$
Commented by ajfour last updated on 06/Apr/20
Fabulous!  Thanks, good question,  great solution.
$${Fabulous}!\:\:{Thanks},\:{good}\:{question}, \\ $$$${great}\:{solution}. \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
you are welcome sir
$${you}\:{are}\:{welcome}\:{sir} \\ $$
Answered by M±th+et£s last updated on 06/Apr/20

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