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Question-93284




Question Number 93284 by naka3546 last updated on 12/May/20
Commented by mr W last updated on 12/May/20
no unique solution!  [EFGH]≥2[ABCD]=4M
$${no}\:{unique}\:{solution}! \\ $$$$\left[{EFGH}\right]\geqslant\mathrm{2}\left[{ABCD}\right]=\mathrm{4}{M} \\ $$
Commented by mr W last updated on 12/May/20
EFGH can be as large as you like,  till ∞.  Every square with vertexes on the   lines L_1 ,L_2 ,L_3 ,L_4  is a solution.
$${EFGH}\:{can}\:{be}\:{as}\:{large}\:{as}\:{you}\:{like}, \\ $$$${till}\:\infty. \\ $$$${Every}\:{square}\:{with}\:{vertexes}\:{on}\:{the}\: \\ $$$${lines}\:{L}_{\mathrm{1}} ,{L}_{\mathrm{2}} ,{L}_{\mathrm{3}} ,{L}_{\mathrm{4}} \:{is}\:{a}\:{solution}. \\ $$
Commented by mr W last updated on 12/May/20
Commented by naka3546 last updated on 12/May/20
thank  you ,  sir.  The  options  are  6M  and  8M
$${thank}\:\:{you}\:,\:\:{sir}. \\ $$$${The}\:\:{options}\:\:{are}\:\:\mathrm{6}{M}\:\:{and}\:\:\mathrm{8}{M}\:\: \\ $$
Commented by mr W last updated on 12/May/20
options are wrong.
$${options}\:{are}\:{wrong}. \\ $$

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