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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

$$findtheareaoftheregularhexagon.$$

Answered by mr W last updated on 13/Feb/23

Commented by mr W last updated on 13/Feb/23

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}\mathit{I}$$$$\left(applyingwhatifoundinQ186924\right)$$$$a=7,b=3,d=5$$$$c^2=\frac{2(7^2+3^2)-5^2}{3}=\frac{91}{3}$$$$s^2=\frac{1}{4}(\frac{91}{3}+5^2+\sqrt{\frac{16\times 7^2\times 3^2-{(3\times \frac{91}{3}-5^2)}^2}{3}}$$$$s^2=\frac{64}{3}\Rightarrow s=\frac{8}{\sqrt{3}}$$$$A_{hexagon}=6\times \frac{\sqrt{3}{s}^2}{4}=32\sqrt{3}$$

Answered by mr W last updated on 13/Feb/23

Commented by mr W last updated on 13/Feb/23

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}$$$$lookattheblueequilateraltriangle$$$$withsidelengthl=\sqrt{3}s.$$$$thedistancesfromapointtothe$$$$vertexesoftheequilateraltriangle$$$$arep=3,q=5,r=7.$$$$asweknowfromQ123040:$$$$l^2=\frac{p^2+q^2+r^2+\sqrt{3\delta }}{2}$$$$with\delta =(p+q+r)(-p+q+r)(p-q+r)(p+q-r)$$$$incurrentcase:$$$$\delta =(3+5+7)(-3+5+7)(3-5+7)(3+5-7)=675$$$$l^2={\left(\sqrt{3}s\right)}^2=\frac{3^2+5^2+7^2+\sqrt{3\times 675}}{2}=64$$$$\Rightarrow s^2=\frac{64}{3}\Rightarrow s=\frac{8}{\sqrt{3}}$$$$A_{hexagon}=6\times \frac{\sqrt{3}{s}^2}{4}=32\sqrt{3}$$

Answered by ajfour last updated on 14/Feb/23

Commented by mr W last updated on 14/Feb/23

$$thanksfortryingsir!$$$$shallwenothave$$$$A=6\times \frac{\sqrt{3}}{4}{\left(s\right)}^2=\frac{3\sqrt{3}}{2}{s}^2?$$

Commented by ajfour last updated on 14/Feb/23

$$Hexagonsidebe\mathit{s}.$$$$circlethroughF(0,0)$$$$x^2+y^2=c^2$$$$throughB(2s\sqrt{3},0)$$$${(x-2s\sqrt{3})}^2+y^2=a^2$$$$subtracting$$$$2s\sqrt{3}(2x-2s\sqrt{3})=c^2-a^2$$$$\Rightarrow x-s\sqrt{3}=\frac{c^2-a^2}{4s\sqrt{3}}...\left(i\right)$$$$or\frac{x}{s}=\sqrt{3}+\frac{c^2-a^2}{4s^2\sqrt{3}}..\left(ii\right)$$$$circlethroughD(s\sqrt{3},\frac{3s}{2})$$$${(x-s\sqrt{3})}^2+{(y-\frac{3s}{2})}^2=b^2$$$$using\left(i\right)$$$${(\frac{y}{s}-\frac{3}{2})}^2=\frac{b^2}{s^2}-\frac{1}{s^4}{\left(\frac{c^2-a^2}{4\sqrt{3}}\right)}^2...\left(I\right)$$$$x^2+y^2=a^2...\left(II\right)$$$${\left(\frac{x}{s}\right)}^2+{\left(\frac{y}{s}\right)}^2=\frac{a^2}{s^2}$$$$using\left(ii\right)$$$${(\sqrt{3}+\frac{c^2-a^2}{4s^2\sqrt{3}})}^2+{\left(\frac{y}{s}\right)}^2=\frac{a^2}{s^2}..\left(III\right)$$$$\left(III\right)-\left(I\right)gives$$$$\frac{3}{2}(\frac{2y}{s}-\frac{3}{2})=\frac{a^2-b^2}{s^2}+{\left(\frac{c^2-a^2}{4s^2\sqrt{3}}\right)}^2-{(\sqrt{3}+\frac{c^2-a^2}{4s^2\sqrt{3}})}^2$$$$\Rightarrow \frac{3}{2}(\frac{3}{2}-\frac{2y}{s})+\frac{a^2-b^2}{s^2}=\sqrt{3}(\sqrt{3}+\frac{c^2-a^2}{2s^2\sqrt{3}})$$$$\Rightarrow \frac{3y}{s}=\frac{a^2-b^2}{s^2}+\frac{a^2-c^2}{2s^2}+\frac{9}{4}-3$$$$\mathit{N}\mathit{o}\mathit{w}$$$$9{\left(\frac{x}{s}\right)}^2+9{\left(\frac{y}{s}\right)}^2=9\left(\frac{a^2}{s^2}\right)$$$$9{(\sqrt{3}+\frac{c^2-a^2}{4s^2\sqrt{3}})}^2+$$$${(\frac{a^2-b^2}{s^2}+\frac{a^2-c^2}{2s^2}+\frac{9}{4}-3)}^2=\frac{9a^2}{s^2}$$$$..$$$$fromhereweget{s}^2$$$$andhenceareaofhexagon$$$$A=6\times \frac{\sqrt{3}}{4}{\left(2s\right)}^2=6\sqrt{3}{s}^2$$$$$$

Commented by ajfour last updated on 14/Feb/23

$$Ihadtakenside=2s$$

Commented by mr W last updated on 14/Feb/23

$$ok.$$

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