Question-49280

Question Number 49280 by mr W last updated on 05/Dec/18

Commented by mr W last updated on 05/Dec/18

$${The}\:{distances}\:{from}\:{its}\:{centroid}\:{to} \\ $$$${its}\:{vertices}\:{are}\:{given}.\:{Find}\:{the}\:{area} \\ $$$${of}\:{the}\:{triangle}. \\ $$

Commented by behi83417@gmail.com last updated on 05/Dec/18

g_a =(2/3)m_a =(1/3)(√(2(b^2 +c^2 )−a^2 ))= =(1/3)(√(2(a^2 +b^2 +c^2 )−3a^2 ))= =(1/3)(√(2×(4/3)Σm_a ^2 −3a^2 ))= =(1/3)(√((8/3).(9/4)Σg_a ^2 −3a^2 ))=(1/3)(√(6Σg_a ^2 −3a^2 )) ⇒3g_a ^2 =2(g_a ^2 +g_b ^2 +g_c ^2 )−a^2 ⇒a^2 =2(g_b ^2 +g_c ^2 )−g_a ^2 ⇒ a=(√(2(g_b ^2 +g_c ^2 )−g_a ^2 )) b=(√(2(g_a ^2 +g_c ^2 )−g_b ^2 )) c=(√(2(g_a ^2 +g_b ^2 )−g_c ^2 )) now we can find area from:Heron′s formula. note:Σm_a ^2 =(3/4)Σa^2 .

$${g}_{{a}} =\frac{\mathrm{2}}{\mathrm{3}}{m}_{{a}} =\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{3}{a}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{2}×\frac{\mathrm{4}}{\mathrm{3}}\Sigma{m}_{{a}} ^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\mathrm{8}}{\mathrm{3}}.\frac{\mathrm{9}}{\mathrm{4}}\Sigma{g}_{{a}} ^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{6}\Sigma{g}_{{a}} ^{\mathrm{2}} −\mathrm{3}{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{3}{g}_{{a}} ^{\mathrm{2}} =\mathrm{2}\left({g}_{{a}} ^{\mathrm{2}} +{g}_{{b}} ^{\mathrm{2}} +{g}_{{c}} ^{\mathrm{2}} \right)−{a}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\mathrm{2}\left({g}_{{b}} ^{\mathrm{2}} +{g}_{{c}} ^{\mathrm{2}} \right)−{g}_{{a}} ^{\mathrm{2}} \Rightarrow \\ $$$$\boldsymbol{\mathrm{a}}=\sqrt{\mathrm{2}\left(\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{b}}} ^{\mathrm{2}} +\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{c}}} ^{\mathrm{2}} \right)−\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{a}}} ^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{b}}=\sqrt{\mathrm{2}\left(\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{a}}} ^{\mathrm{2}} +\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{c}}} ^{\mathrm{2}} \right)−\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{b}}} ^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{c}}=\sqrt{\mathrm{2}\left(\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{a}}} ^{\mathrm{2}} +\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{b}}} ^{\mathrm{2}} \right)−\boldsymbol{\mathrm{g}}_{\boldsymbol{\mathrm{c}}} ^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{now}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{from}}:{Heron}'{s}\:\boldsymbol{\mathrm{formula}}. \\ $$$$\boldsymbol{\mathrm{note}}:\Sigma{m}_{{a}} ^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}}\Sigma{a}^{\mathrm{2}} . \\ $$

Commented by mr W last updated on 05/Dec/18

$${thanks}\:{sir}! \\ $$

Answered by MJS last updated on 05/Dec/18

A= ((0),(0) ) B= ((c),(0) ) C= ((((−a^2 +b^2 +c^2 )/(2c))),(((δ(a, b, c))/(2c))) ) δ(a, b, c)=(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c))) G= ((((−a^2 +b^2 +3c^2 )/(6c))),(((δ(a, b, c))/(6c))) ) g_a ^2 =((−a^2 +2b^2 +2c^2 )/9) g_b ^2 =((2a^2 −b^2 +2c^2 )/9) g_c ^2 =((2a^2 +2b^2 −c^2 )/9) ⇒ a^2 =−g_a ^2 +2g_b ^2 +2g_c ^2 b^2 =2g_a ^2 −g_b ^2 +2g_c ^2 c^2 =2g_a ^2 +2g_b ^2 −g_c ^2 ⇒ area(a, b, c)=((δ(a, b, c))/4)=((3δ(g_a , g_b , g_c ))/4)

$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{\frac{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{c}}}\\{\frac{\delta\left({a},\:{b},\:{c}\right)}{\mathrm{2}{c}}}\end{pmatrix} \\ $$$$\delta\left({a},\:{b},\:{c}\right)=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)} \\ $$$${G}=\begin{pmatrix}{\frac{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{6}{c}}}\\{\frac{\delta\left({a},\:{b},\:{c}\right)}{\mathrm{6}{c}}}\end{pmatrix} \\ $$$${g}_{{a}} ^{\mathrm{2}} =\frac{−{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{9}} \\ $$$${g}_{{b}} ^{\mathrm{2}} =\frac{\mathrm{2}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{9}} \\ $$$${g}_{{c}} ^{\mathrm{2}} =\frac{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\Rightarrow \\ $$$${a}^{\mathrm{2}} =−{g}_{{a}} ^{\mathrm{2}} +\mathrm{2}{g}_{{b}} ^{\mathrm{2}} +\mathrm{2}{g}_{{c}} ^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} =\mathrm{2}{g}_{{a}} ^{\mathrm{2}} −{g}_{{b}} ^{\mathrm{2}} +\mathrm{2}{g}_{{c}} ^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} =\mathrm{2}{g}_{{a}} ^{\mathrm{2}} +\mathrm{2}{g}_{{b}} ^{\mathrm{2}} −{g}_{{c}} ^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{area}\left({a},\:{b},\:{c}\right)=\frac{\delta\left({a},\:{b},\:{c}\right)}{\mathrm{4}}=\frac{\mathrm{3}\delta\left({g}_{{a}} ,\:{g}_{{b}} ,\:{g}_{{c}} \right)}{\mathrm{4}} \\ $$

Commented by mr W last updated on 05/Dec/18

$${thank}\:{you}\:{sir}! \\ $$$${the}\:{last}\:{line}\:{is}\:{new}\:{to}\:{me},\:{very}\:{nice}! \\ $$

Commented by MJS last updated on 05/Dec/18

$$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{sentence} \\ $$

Commented by mr W last updated on 05/Dec/18

Commented by mr W last updated on 05/Dec/18

Area of hexagon=2×Area of ΔABC Area of hexagon=6×Area of triangle with sides g_a ,g_b ,g_c ⇒Area of ΔABC=3×Area of triangle with sides g_a ,g_b ,g_c

$${Area}\:{of}\:{hexagon}=\mathrm{2}×{Area}\:{of}\:\Delta{ABC} \\ $$$${Area}\:{of}\:{hexagon}=\mathrm{6}×{Area}\:{of}\:{triangle}\:{with}\:{sides}\:{g}_{{a}} ,{g}_{{b}} ,{g}_{{c}} \\ $$$$\Rightarrow{Area}\:{of}\:\Delta{ABC}=\mathrm{3}×{Area}\:{of}\:{triangle}\:{with}\:{sides}\:{g}_{{a}} ,{g}_{{b}} ,{g}_{{c}} \\ $$

Commented by mr W last updated on 05/Dec/18

$${Now}\:{I}\:{understand}\:{it}\:{geometrically}. \\ $$

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