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Question Number 176100 by mnjuly1970 last updated on 12/Sep/22

Q : in AB^Δ C : A = 90^o and , m_( b) ^2 + m_( c) ^( 2) = 100. find the value of ” a ” . note : ⟨ m_( a) := the median corresponding to ” A ” . −−− m.n−−−

$$ \\ $$$$\:\:\:\:\:{Q}\:: \\ $$$$\:\:\:\:\:{in}\:\:{A}\overset{\Delta} {{B}C}\::\:{A}\:=\:\mathrm{90}^{\mathrm{o}} \:\:\:{and}\:,\:{m}_{\:{b}} ^{\mathrm{2}} \:+\:{m}_{\:{c}} ^{\:\mathrm{2}} \:=\:\mathrm{100}.\:\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{find}\:\:{the}\:\:{value}\:{of}\:\:\:''\:\:\:{a}\:\:\:''\:. \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{note}\::\:\:\langle\:{m}_{\:{a}} \::=\:{the}\:{median}\:{corresponding}\:\:{to}\:''\:{A}\:''\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−−−\:\:\boldsymbol{{m}}.\boldsymbol{{n}}−−− \\ $$

Answered by Rasheed.Sindhi last updated on 12/Sep/22

• a^2 =b^2 +c^2 • b^2 +((c/2))^2 =m_c ^2 • c^2 +((b/2))^2 =m_b ^2 m_( b) ^2 + m_( c) ^( 2) ={c^2 +((b/2))^2 }+{b^2 +((c/2))^2 }=100 c^2 +(b^2 /4)+b^2 +(c^2 /4)=100 5b^2 +5c^2 =400 b^2 +c^2 =80 a^2 =80 a=(√(80)) =4(√5)

$$\bullet\:{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\bullet\:{b}^{\mathrm{2}} +\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} ={m}_{{c}} ^{\mathrm{2}} \\ $$$$\bullet\:{c}^{\mathrm{2}} +\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} ={m}_{{b}} ^{\mathrm{2}} \\ $$$${m}_{\:{b}} ^{\mathrm{2}} \:+\:{m}_{\:{c}} ^{\:\mathrm{2}} \\ $$$$\:\:\:\:\:=\left\{{c}^{\mathrm{2}} +\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} \right\}+\left\{{b}^{\mathrm{2}} +\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} \right\}=\mathrm{100} \\ $$$$\:\:\:\:\:\:\:{c}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+{b}^{\mathrm{2}} +\frac{{c}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{100} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{5}{b}^{\mathrm{2}} +\mathrm{5}{c}^{\mathrm{2}} =\mathrm{400} \\ $$$$\:\:\:\:\:\:\:\:\:\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{80} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} =\mathrm{80} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}=\sqrt{\mathrm{80}}\:=\mathrm{4}\sqrt{\mathrm{5}}\: \\ $$

Commented by mnjuly1970 last updated on 12/Sep/22

grateful sir rasheed

$${grateful}\:{sir}\:{rasheed} \\ $$

Answered by behi834171 last updated on 13/Sep/22

2m_b =a^2 +c^2 −(b^2 /2) 2m_c =a^2 +b^2 −(c^2 /2)⇒2(m_b ^2 +m_c ^2 )=2a^2 +(a^2 /2)⇒ m_b ^2 +m_c ^2 =((5a^2 )/4)=5((a/2))^2 =5m_a ^2 ⇒200=((5a^2 )/2)⇒a=(√(80))=4(√5) . ■

$$\mathrm{2}{m}_{{b}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{2}{m}_{{c}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow\mathrm{2}\left({m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} \right)=\mathrm{2}{a}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow \\ $$$$\boldsymbol{{m}}_{\boldsymbol{{b}}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\boldsymbol{{c}}} ^{\mathrm{2}} =\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{5}\left(\frac{\boldsymbol{{a}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{5}\boldsymbol{{m}}_{\boldsymbol{{a}}} ^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{200}=\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow\boldsymbol{{a}}=\sqrt{\mathrm{80}}=\mathrm{4}\sqrt{\mathrm{5}}\:\:.\:\blacksquare \\ $$

Commented by mnjuly1970 last updated on 13/Sep/22

mamnoon ostad

$${mamnoon}\:{ostad} \\ $$

Commented by behi834171 last updated on 13/Sep/22

[note: when:∡A=90^• ⇒m_a =(a/2),m_b ^2 +m_c ^2 =5m_a ^2 ⇒m_a ^2 =((m_b ^2 +m_c ^2 )/5)=((100)/5)=20 ⇒m_a =2(√5)⇒a=2m_a =4(√5) .]

$$\left[{note}:\right. \\ $$$${when}:\measuredangle{A}=\mathrm{90}^{\bullet} \Rightarrow\boldsymbol{{m}}_{\boldsymbol{{a}}} =\frac{\boldsymbol{{a}}}{\mathrm{2}},\boldsymbol{{m}}_{\boldsymbol{{b}}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\boldsymbol{{c}}} ^{\mathrm{2}} =\mathrm{5}\boldsymbol{{m}}_{\boldsymbol{{a}}} ^{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{{m}}_{\boldsymbol{{a}}} ^{\mathrm{2}} =\frac{\boldsymbol{{m}}_{\boldsymbol{{b}}} ^{\mathrm{2}} +\boldsymbol{{m}}_{\boldsymbol{{c}}} ^{\mathrm{2}} }{\mathrm{5}}=\frac{\mathrm{100}}{\mathrm{5}}=\mathrm{20} \\ $$$$\left.\Rightarrow\boldsymbol{{m}}_{\boldsymbol{{a}}} =\mathrm{2}\sqrt{\mathrm{5}}\Rightarrow\boldsymbol{{a}}=\mathrm{2}\boldsymbol{{m}}_{\boldsymbol{{a}}} =\mathrm{4}\sqrt{\mathrm{5}}\:.\right] \\ $$

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