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Question Number 152204 by cherokeesay last updated on 26/Aug/21

Commented by Paradoxical last updated on 26/Aug/21

Commented by cherokeesay last updated on 26/Aug/21

Nice ! thank you !

$${Nice}\:! \\ $$$${thank}\:{you}\:! \\ $$

Answered by Rasheed.Sindhi last updated on 26/Aug/21

^(−) Length=x ,Width=y Small trianvles are similar: (∵ ∢s are congruent) ∴ (x/3)=(4/y)⇒xy=12⇒A=12 sq units. ^(−)

$$\underline{\overline {\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}} \\ $$$${Length}={x}\:,{Width}={y} \\ $$$${Small}\:{trianvles}\:{are}\:{similar}: \\ $$$$\left(\because\:\sphericalangle{s}\:{are}\:{congruent}\right) \\ $$$$\therefore\:\:\frac{{x}}{\mathrm{3}}=\frac{\mathrm{4}}{{y}}\Rightarrow{xy}=\mathrm{12}\Rightarrow{A}=\mathrm{12}\:\mathrm{sq}\:\mathrm{units}. \\ $$$$\:\:\underline{\overline {\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}} \\ $$

Commented by cherokeesay last updated on 26/Aug/21

very clever ! thank you !

$${very}\:{clever}\:! \\ $$$${thank}\:{you}\:! \\ $$

Answered by Paradoxical last updated on 26/Aug/21

Commented by Rasheed.Sindhi last updated on 27/Aug/21

paradoxical sir, Why have you put two identical versions of your answer? :)

$${paradoxical}\:{sir},\:{Why}\:{have}\:{you}\:{put} \\ $$$${two}\:\boldsymbol{{identical}}\:{versions}\:{of}\:{your} \\ $$$$\left.{answer}?\::\right) \\ $$

Answered by Rasheed.Sindhi last updated on 27/Aug/21

⟨⟨Through Hypotenuses⟩⟩ Sum of Hypotenuses of small triangles =Hypotenuse of big triangle Length=x,Width=y (√(4^2 +x^2 ))+(√(y^2 +3^2 ))=(√((4+y)^2 +(x+3)^2 )) 16+x^2 +y^2 +9+2(√(4^2 +x^2 ))(√(y^2 +3^2 )) =16+8y+y^2 +x^2 +6x+9 2(√(4^2 +x^2 ))(√(y^2 +3^2 ))=8y+6x (√(4^2 +x^2 ))(√(y^2 +3^2 ))=4y+3x (16+x^2 )(y^2 +9)=16y^2 +24xy+9x^2 x^2 y^2 +9x^2 +16y^2 +144=16y^2 +24xy+9x^2 (xy)^2 −24(xy)+144=0 (xy−12)^2 =0 xy=12

$$\langle\langle\mathcal{T}{hrough}\:\mathcal{H}{ypotenuses}\rangle\rangle \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{Hypotenuses}\:\mathrm{of}\:\mathrm{small}\:\mathrm{triangles} \\ $$$$\:\:\:\:\:\:=\mathrm{Hypotenuse}\:\mathrm{of}\:\mathrm{big}\:\mathrm{triangle} \\ $$$${Length}={x},{Width}={y} \\ $$$$\sqrt{\mathrm{4}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\sqrt{{y}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\sqrt{\left(\mathrm{4}+{y}\right)^{\mathrm{2}} +\left({x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{16}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{9}+\mathrm{2}\sqrt{\mathrm{4}^{\mathrm{2}} +{x}^{\mathrm{2}} }\sqrt{{y}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{16}+\mathrm{8}{y}+{y}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9} \\ $$$$\mathrm{2}\sqrt{\mathrm{4}^{\mathrm{2}} +{x}^{\mathrm{2}} }\sqrt{{y}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\mathrm{8}{y}+\mathrm{6}{x} \\ $$$$\sqrt{\mathrm{4}^{\mathrm{2}} +{x}^{\mathrm{2}} }\sqrt{{y}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\mathrm{4}{y}+\mathrm{3}{x} \\ $$$$\left(\mathrm{16}+{x}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} +\mathrm{9}\right)=\mathrm{16}{y}^{\mathrm{2}} +\mathrm{24}{xy}+\mathrm{9}{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{9}{x}^{\mathrm{2}} +\mathrm{16}{y}^{\mathrm{2}} +\mathrm{144}=\mathrm{16}{y}^{\mathrm{2}} +\mathrm{24}{xy}+\mathrm{9}{x}^{\mathrm{2}} \\ $$$$\left({xy}\right)^{\mathrm{2}} −\mathrm{24}\left({xy}\right)+\mathrm{144}=\mathrm{0} \\ $$$$\left({xy}−\mathrm{12}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${xy}=\mathrm{12} \\ $$

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