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Question-187066

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Question Number 187066 by mr W last updated on 13/Feb/23
Commented by mr W last updated on 13/Feb/23
find the area of the regular hexagon.
$${find}\:{the}\:{area}\:{of}\:{the}\:{regular}\:{hexagon}. \\ $$
Answered by mr W last updated on 13/Feb/23
Commented by mr W last updated on 13/Feb/23
Method II (applying what i found in Q186924) a=7, b=3, d=5 c^2 =((2(7^2 +3^2 )−5^2 )/3)=((91)/3) s^2 =(1/4)(((91)/3)+5^2 +(√((16×7^2 ×3^2 −(3×((91)/3)−5^2 )^2 )/3)) s^2 =((64)/3) ⇒s=(8/( (√3))) A_(hexagon) =6×(((√3)s^2 )/4)=32(√3)
$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$$\left({applying}\:{what}\:{i}\:{found}\:{in}\:{Q}\mathrm{186924}\right) \\ $$$${a}=\mathrm{7},\:{b}=\mathrm{3},\:{d}=\mathrm{5} \\ $$$${c}^{\mathrm{2}} =\frac{\mathrm{2}\left(\mathrm{7}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)−\mathrm{5}^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{91}}{\mathrm{3}} \\ $$$${s}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{91}}{\mathrm{3}}+\mathrm{5}^{\mathrm{2}} +\sqrt{\frac{\mathrm{16}×\mathrm{7}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} −\left(\mathrm{3}×\frac{\mathrm{91}}{\mathrm{3}}−\mathrm{5}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{3}}}\right. \\ $$$${s}^{\mathrm{2}} =\frac{\mathrm{64}}{\mathrm{3}}\:\Rightarrow{s}=\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}} \\ $$$${A}_{{hexagon}} =\mathrm{6}×\frac{\sqrt{\mathrm{3}}{s}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{32}\sqrt{\mathrm{3}} \\ $$
Answered by mr W last updated on 13/Feb/23
Commented by mr W last updated on 13/Feb/23
Method I look at the blue equilateral triangle with side length l=(√3)s. the distances from a point to the vertexes of the equilateral triangle are p=3, q=5, r=7. as we know from Q123040: l^2 =((p^2 +q^2 +r^2 +(√(3δ)))/2) with δ=(p+q+r)(−p+q+r)(p−q+r)(p+q−r) in current case: δ=(3+5+7)(−3+5+7)(3−5+7)(3+5−7)=675 l^2 =((√3)s)^2 =((3^2 +5^2 +7^2 +(√(3×675)))/2)=64 ⇒s^2 =((64)/3) ⇒s=(8/( (√3))) A_(hexagon) =6×(((√3)s^2 )/4)=32(√3)
$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$${look}\:{at}\:{the}\:{blue}\:{equilateral}\:{triangle} \\ $$$${with}\:{side}\:{length}\:{l}=\sqrt{\mathrm{3}}{s}. \\ $$$${the}\:{distances}\:{from}\:{a}\:{point}\:{to}\:{the} \\ $$$${vertexes}\:{of}\:{the}\:{equilateral}\:{triangle} \\ $$$${are}\:{p}=\mathrm{3},\:{q}=\mathrm{5},\:{r}=\mathrm{7}. \\ $$$${as}\:{we}\:{know}\:{from}\:{Q}\mathrm{123040}: \\ $$$${l}^{\mathrm{2}} =\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} +\sqrt{\mathrm{3}\delta}}{\mathrm{2}} \\ $$$${with}\:\delta=\left({p}+{q}+{r}\right)\left(−{p}+{q}+{r}\right)\left({p}−{q}+{r}\right)\left({p}+{q}−{r}\right) \\ $$$${in}\:{current}\:{case}: \\ $$$$\delta=\left(\mathrm{3}+\mathrm{5}+\mathrm{7}\right)\left(−\mathrm{3}+\mathrm{5}+\mathrm{7}\right)\left(\mathrm{3}−\mathrm{5}+\mathrm{7}\right)\left(\mathrm{3}+\mathrm{5}−\mathrm{7}\right)=\mathrm{675} \\ $$$${l}^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}{s}\right)^{\mathrm{2}} =\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} +\sqrt{\mathrm{3}×\mathrm{675}}}{\mathrm{2}}=\mathrm{64} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{\mathrm{64}}{\mathrm{3}}\:\Rightarrow{s}=\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}} \\ $$$${A}_{{hexagon}} =\mathrm{6}×\frac{\sqrt{\mathrm{3}}{s}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{32}\sqrt{\mathrm{3}} \\ $$
Answered by ajfour last updated on 14/Feb/23
Commented by mr W last updated on 14/Feb/23
thanks for trying sir! shall we not have A=6×((√3)/4)(s)^2 =((3(√3))/2)s^2 ?
$${thanks}\:{for}\:{trying}\:{sir}! \\ $$$${shall}\:{we}\:{not}\:{have} \\ $$$${A}=\mathrm{6}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({s}\right)^{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}{s}^{\mathrm{2}} \:\:? \\ $$
Commented by ajfour last updated on 14/Feb/23
Hexagon side be s. circle through F(0,0) x^2 +y^2 =c^2 through B(2s(√3), 0) (x−2s(√3))^2 +y^2 =a^2 subtracting 2s(√3)(2x−2s(√3))=c^2 −a^2 ⇒ x−s(√3)=((c^2 −a^2 )/(4s(√3))) ...(i) or (x/s)=(√3)+((c^2 −a^2 )/(4s^2 (√3))) ..(ii) circle through D(s(√3), ((3s)/2)) (x−s(√3))^2 +(y−((3s)/2))^2 =b^2 using (i) ((y/s)−(3/2))^2 =(b^2 /s^2 )−(1/s^4 )(((c^2 −a^2 )/(4(√3))))^2 ...(I) x^2 +y^2 =a^2 ...(II) ((x/s))^2 +((y/s))^2 =(a^2 /s^2 ) using (ii) ((√3)+((c^2 −a^2 )/(4s^2 (√3))))^2 +((y/s))^2 =(a^2 /s^2 ) ..(III) (III)−(I) gives (3/2)(((2y)/s)−(3/2))=((a^2 −b^2 )/s^2 )+(((c^2 −a^2 )/(4s^2 (√3))))^2 −((√3)+((c^2 −a^2 )/(4s^2 (√3))))^2 ⇒ (3/2)((3/2)−((2y)/s))+((a^2 −b^2 )/s^2 )=(√3)((√3)+((c^2 −a^2 )/(2s^2 (√3)))) ⇒ ((3y)/s)=((a^2 −b^2 )/s^2 )+((a^2 −c^2 )/(2s^2 ))+(9/4)−3 Now 9((x/s))^2 +9((y/s))^2 =9((a^2 /s^2 )) 9((√3)+((c^2 −a^2 )/(4s^2 (√3))))^2 + (((a^2 −b^2 )/s^2 )+((a^2 −c^2 )/(2s^2 ))+(9/4)−3)^2 =((9a^2 )/s^2 ) .. from here we get s^2 and hence area of hexagon A=6×((√3)/4)(2s)^2 =6(√3)s^2
$${Hexagon}\:{side}\:{be}\:\boldsymbol{{s}}. \\ $$$${circle}\:{through}\:{F}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$${through}\:{B}\left(\mathrm{2}{s}\sqrt{\mathrm{3}},\:\mathrm{0}\right) \\ $$$$\left({x}−\mathrm{2}{s}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${subtracting} \\ $$$$\mathrm{2}{s}\sqrt{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{2}{s}\sqrt{\mathrm{3}}\right)={c}^{\mathrm{2}} −{a}^{\mathrm{2}} \:\: \\ $$$$\Rightarrow\:\:{x}−{s}\sqrt{\mathrm{3}}=\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}\sqrt{\mathrm{3}}}\:\:\:\:…\left({i}\right) \\ $$$${or}\:\:\frac{{x}}{{s}}=\sqrt{\mathrm{3}}+\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}^{\mathrm{2}} \sqrt{\mathrm{3}}}\:\:\:\:..\left({ii}\right) \\ $$$${circle}\:{through}\:{D}\left({s}\sqrt{\mathrm{3}},\:\frac{\mathrm{3}{s}}{\mathrm{2}}\right) \\ $$$$\left({x}−{s}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{3}{s}}{\mathrm{2}}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$${using}\:\left({i}\right) \\ $$$$\left(\frac{{y}}{{s}}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{s}^{\mathrm{2}} }−\frac{\mathrm{1}}{{s}^{\mathrm{4}} }\left(\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \:\:\:…\left({I}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({II}\right) \\ $$$$\left(\frac{{x}}{{s}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{s}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{{s}^{\mathrm{2}} } \\ $$$${using}\:\:\left({ii}\right) \\ $$$$\left(\sqrt{\mathrm{3}}+\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}^{\mathrm{2}} \sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{s}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{{s}^{\mathrm{2}} }\:\:\:\:..\left({III}\right) \\ $$$$\left({III}\right)−\left({I}\right)\:\:{gives} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{2}{y}}{{s}}−\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{s}^{\mathrm{2}} }+\left(\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}^{\mathrm{2}} \sqrt{\mathrm{3}}}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}+\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}^{\mathrm{2}} \sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{2}{y}}{{s}}\right)+\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{s}^{\mathrm{2}} }=\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{s}^{\mathrm{2}} \sqrt{\mathrm{3}}}\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{3}{y}}{{s}}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{s}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{s}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{3} \\ $$$$\boldsymbol{{Now}}\:\:\: \\ $$$$\mathrm{9}\left(\frac{{x}}{{s}}\right)^{\mathrm{2}} +\mathrm{9}\left(\frac{{y}}{{s}}\right)^{\mathrm{2}} =\mathrm{9}\left(\frac{{a}^{\mathrm{2}} }{{s}^{\mathrm{2}} }\right) \\ $$$$\mathrm{9}\left(\sqrt{\mathrm{3}}+\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}^{\mathrm{2}} \sqrt{\mathrm{3}}}\right)^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\left(\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{s}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{s}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{3}\right)^{\mathrm{2}} =\frac{\mathrm{9}{a}^{\mathrm{2}} }{{s}^{\mathrm{2}} } \\ $$$$.. \\ $$$${from}\:{here}\:{we}\:{get}\:{s}^{\mathrm{2}} \\ $$$${and}\:{hence}\:{area}\:{of}\:{hexagon} \\ $$$${A}=\mathrm{6}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{2}{s}\right)^{\mathrm{2}} =\mathrm{6}\sqrt{\mathrm{3}}{s}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by ajfour last updated on 14/Feb/23
I had taken side=2s
$${I}\:{had}\:{taken}\:{side}=\mathrm{2}{s} \\ $$
Commented by mr W last updated on 14/Feb/23
ok.
$${ok}. \\ $$

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