Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 25605 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17

Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17

$$\mathrm{from}\mathrm{midpoints}\mathrm{of}\mathrm{sides}\mathrm{draw}\mathrm{perpendicular}$$$$\mathrm{lines}\mathrm{to}\mathrm{opposite}\mathrm{sides}.\mathrm{find}\mathrm{the}\mathrm{ratio}\mathrm{of}$$$$\mathrm{area}\mathrm{of}\mathrm{inner}\mathrm{hexagon}\mathrm{to}\mathrm{area}\mathrm{of}:\mathrm{A}\stackrel{△}{\mathrm{B}C}.$$

Commented by ajfour last updated on 12/Dec/17

Commented by ajfour last updated on 12/Dec/17

$$△ABC=\frac{1}{2}\left(2a\right)\left(2c\right)=2ac$$$$△DEF=\frac{1}{4}(△ABC)=\frac{ac}{2}$$$$△DGF=\frac{1}{2}\times FM\times (MD-GM)$$$$=\frac{1}{2}\times b\times (c-\frac{b}{\tan C})=\frac{b}{2}[c-\frac{b(a-b)}{c}]$$$$=\frac{\mathit{b}({\mathit{c}}^2+{\mathit{b}}^2-\mathit{a}\mathit{b})}{2\mathit{c}}.$$$$△FEB=\frac{1}{2}\times NF\times (EN-IN)$$$$=\frac{1}{2}(a-b)[c-\frac{a-b}{\tan B}]$$$$=\frac{1}{2}(a-b)[c-\frac{(a-b)b}{c}]$$$$=\frac{(\mathit{a}-\mathit{b})({\mathit{c}}^2+{\mathit{b}}^2-\mathit{a}\mathit{b})}{2\mathit{c}}.$$$$andy(\tan B+\tan C)=a$$$$\Rightarrow y=\frac{a}{(\frac{c}{b}+\frac{c}{a-b})}=\frac{b(a-b)}{c}$$$$△HED=\frac{1}{2}\times \mathit{a}\times \mathit{y}$$$$=\frac{ab(a-b)}{2c}$$$$Area\left(hexagon\right)=Ar(△DEF)+$$$$Ar(△DGF)+Ar(△FEB)+$$$$Ar(△HED)$$$$=\frac{ac}{2}+\frac{b}{2c}(c^2+b^2-ab)+$$$$\frac{(a-b)(c^2+b^2-ab)}{2c}+\frac{ab(a-b)}{2c}$$$$=\frac{ac}{2}+\frac{(c^2+b^2-ab)a}{2c}+\frac{(ab-b^2)a}{2c}$$$$=\frac{a}{2c}(c^2+c^2)=ac$$$$so\frac{Ar\left(hexagon\right)}{Ar(△ABC)}=\frac{\mathit{a}\mathit{c}}{2\mathit{a}\mathit{c}}=\frac{1}{2}.$$

Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{mr}\mathrm{Ajfour}.$$$$\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{elegant}\mathrm{approach}!$$

Answered by mrW1 last updated on 12/Dec/17

Commented by mrW1 last updated on 12/Dec/17

$$Anotherway:$$$$\mathrm{\Delta }DEF=\frac{1}{4}\mathrm{\Delta }ABC$$$$\mathrm{\Delta }DIE=\mathrm{\Delta }AGF=\mathrm{\Delta }FHC$$$$\mathrm{\Delta }FGD=\mathrm{\Delta }CHE=\mathrm{\Delta }EIB$$$$\mathrm{\Delta }EHF=\mathrm{\Delta }BID=\mathrm{\Delta }DGA$$$$\Rightarrow \mathrm{\Delta }DIE+\mathrm{\Delta }FGD+\mathrm{\Delta }EHF=\mathrm{\Delta }BDE=\frac{1}{4}\mathrm{\Delta }ABC$$$$A_{HEXAGON}=\mathrm{\Delta }DEF+\mathrm{\Delta }DIE+\mathrm{\Delta }FGD+\mathrm{\Delta }EHF$$$$=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\mathrm{\Delta }ABC$$

Commented by ajfour last updated on 12/Dec/17

$$elegantshadingSir!$$

Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17

$$\mathrm{Hi}\mathrm{dear}\mathrm{master}.\mathrm{as}\mathrm{mr}\mathrm{Ajfour}\mathrm{said},\mathrm{it}\mathrm{is}$$$$\mathrm{a}\mathrm{beautiful}\mathrm{and}\mathrm{smart}\mathrm{solution}.\mathrm{thanks}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com