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Question-207683




Question Number 207683 by efronzo1 last updated on 22/May/24
Answered by mr W last updated on 23/May/24
Commented by mr W last updated on 22/May/24
MK=(√(9^2 +(6−3)^2 ))=3(√(10))  MN=(√(9^2 +(6+3)^2 ))=9(√2)  sin α=(9/( 9(√2)))=(1/( (√2)))  R=((MK)/(2 sin α))=((3(√(10)))/(2×(1/( (√2)))))=3(√5)  area of square =(2R)^2 =(6(√5))^2 =180
$${MK}=\sqrt{\mathrm{9}^{\mathrm{2}} +\left(\mathrm{6}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{10}} \\ $$$${MN}=\sqrt{\mathrm{9}^{\mathrm{2}} +\left(\mathrm{6}+\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{9}\sqrt{\mathrm{2}} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{9}}{\:\mathrm{9}\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${R}=\frac{{MK}}{\mathrm{2}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}=\mathrm{3}\sqrt{\mathrm{5}} \\ $$$${area}\:{of}\:{square}\:=\left(\mathrm{2}{R}\right)^{\mathrm{2}} =\left(\mathrm{6}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{180} \\ $$
Commented by Tawa11 last updated on 21/Jun/24
weldone sir
$$\mathrm{weldone}\:\mathrm{sir} \\ $$
Answered by efronzo1 last updated on 23/May/24
 Let  { ((CH= a)),((AE= b)) :} then  { ((36=a(9+b))),((9=b(9+a))) :}     { ((a = b+3)),(((b+12)b−9=0)) :} ⇒b = 3(√5)−6   and a = 3(√5)−3      S[ABCD] = (9+a+b)^2  = (6(√5) )^2 =180 cm^2
$$\:\mathrm{Let}\:\begin{cases}{\mathrm{CH}=\:{a}}\\{{A}\mathrm{E}=\:\mathrm{b}}\end{cases}\:\mathrm{then}\:\begin{cases}{\mathrm{36}={a}\left(\mathrm{9}+{b}\right)}\\{\mathrm{9}={b}\left(\mathrm{9}+{a}\right)}\end{cases} \\ $$$$\:\:\begin{cases}{{a}\:=\:{b}+\mathrm{3}}\\{\left({b}+\mathrm{12}\right){b}−\mathrm{9}=\mathrm{0}}\end{cases}\:\Rightarrow{b}\:=\:\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{6} \\ $$$$\:\mathrm{and}\:{a}\:=\:\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{3}\: \\ $$$$\:\:\:\mathrm{S}\left[\mathrm{ABCD}\right]\:=\:\left(\mathrm{9}+{a}+{b}\right)^{\mathrm{2}} \:=\:\left(\mathrm{6}\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} =\mathrm{180}\:{cm}^{\mathrm{2}} \\ $$

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