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Question-73221




Question Number 73221 by aliesam last updated on 08/Nov/19
Answered by MJS last updated on 08/Nov/19
A= ((0),(0) )  B= ((a),(0) )  C= ((a),(a) )  D= ((0),(a) )                     E= ((b),(0) )  F= ((b),(b) )  G= ((0),(b) )  T= ((((a+b)/2)),(0) )  I= ((((a+b)/2)),(((a−b)/2)) )  M= ((b),(((a−b)/2)) )  N= (((((√3)/4)(a+b)+b)),(((3a−b)/4)) )  L= ((b),(a) )  radius of incircle of ETIM=(r/2)=((a−b)/4)  radius of incircle of MNL=((√3)/(12))(a+b)  ((a−b)/4)=((√3)/(12))(a+b) ⇒ a=(2+(√3))b ⇒ r=((1+(√3))/2)b  ((ab)/(a−b))=(((2+(√3))b^2 )/((2+(√3))b−b))=((1+(√3))/2)b
$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{a}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{a}}\\{{a}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{E}=\begin{pmatrix}{{b}}\\{\mathrm{0}}\end{pmatrix}\:\:{F}=\begin{pmatrix}{{b}}\\{{b}}\end{pmatrix}\:\:{G}=\begin{pmatrix}{\mathrm{0}}\\{{b}}\end{pmatrix} \\ $$$${T}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\:{I}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\frac{{a}−{b}}{\mathrm{2}}}\end{pmatrix} \\ $$$${M}=\begin{pmatrix}{{b}}\\{\frac{{a}−{b}}{\mathrm{2}}}\end{pmatrix}\:\:{N}=\begin{pmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({a}+{b}\right)+{b}}\\{\frac{\mathrm{3}{a}−{b}}{\mathrm{4}}}\end{pmatrix}\:\:{L}=\begin{pmatrix}{{b}}\\{{a}}\end{pmatrix} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{incircle}\:\mathrm{of}\:{ETIM}=\frac{{r}}{\mathrm{2}}=\frac{{a}−{b}}{\mathrm{4}} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{incircle}\:\mathrm{of}\:{MNL}=\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\left({a}+{b}\right) \\ $$$$\frac{{a}−{b}}{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\left({a}+{b}\right)\:\Rightarrow\:{a}=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){b}\:\Rightarrow\:{r}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}{b} \\ $$$$\frac{{ab}}{{a}−{b}}=\frac{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){b}^{\mathrm{2}} }{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){b}−{b}}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}{b} \\ $$

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