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Question Number 207980 by mnjuly1970 last updated on 01/Jun/24

In AB^Δ C : B= 90^( o) BB′ ⊥ CC′ ( BB′ and CC′ are medians) ⇒ (m_c /(m_a )) = ? note: ∣ CC′ ∣ = m_c , ∣AA′∣= m_a

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{In}\:\:\:{A}\overset{\Delta} {{B}C}\::\:\:{B}=\:\mathrm{90}^{\:\mathrm{o}} \: \\ $$$$\:\mathrm{BB}'\:\:\bot\:\mathrm{CC}'\:\left(\:\mathrm{BB}'\:{and}\:\mathrm{CC}'\:{are}\:{medians}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\frac{\boldsymbol{{m}}_{\boldsymbol{{c}}} }{\boldsymbol{{m}}_{\boldsymbol{{a}}} \:}\:=\:? \\ $$$$\boldsymbol{{note}}:\:\:\mid\:\mathrm{CC}'\:\mid\:=\:\boldsymbol{{m}}_{\boldsymbol{{c}}} \:\:,\:\mid\mathrm{AA}'\mid=\:\boldsymbol{{m}}_{\boldsymbol{{a}}} \\ $$

Answered by mr W last updated on 02/Jun/24

Commented by mr W last updated on 02/Jun/24

B′G=(m_b /3)=((√(2(a^2 +c^2 )−b^2 ))/6) CG=((2m_c )/3)=((√(2(a^2 +b^2 )−c^2 ))/3) B′C=(b/2) (((√(2(a^2 +c^2 )−b^2 ))/6))^2 +(((√(2(a^2 +b^2 )−c^2 ))/3))^2 =((b/2))^2 ((2(a^2 +c^2 )−b^2 )/(36))+((2(a^2 +b^2 )−c^2 )/9)=(b^2 /4) ⇒5a^2 =b^2 +c^2 b^2 =a^2 +c^2 ⇒2a^2 =c^2 (m_c /m_a )=(√((2(a^2 +b^2 )−c^2 )/(2(b^2 +c^2 )−a^2 ))) =(√((2(2a^2 +2a^2 )−2a^2 )/(2(a^2 +4a^2 )−a^2 )))=(√(2/3)) ✓

$${B}'{G}=\frac{{m}_{{b}} }{\mathrm{3}}=\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} }}{\mathrm{6}} \\ $$$${CG}=\frac{\mathrm{2}{m}_{{c}} }{\mathrm{3}}=\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} }}{\mathrm{3}} \\ $$$${B}'{C}=\frac{{b}}{\mathrm{2}} \\ $$$$\left(\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} }}{\mathrm{6}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} }}{\mathrm{3}}\right)^{\mathrm{2}} =\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} }{\mathrm{36}}+\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} }{\mathrm{9}}=\frac{{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{5}{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{a}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\frac{{m}_{{c}} }{{m}_{{a}} }=\sqrt{\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} }{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:=\sqrt{\frac{\mathrm{2}\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} \right)−\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }}=\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\:\checkmark \\ $$

Commented by efronzo1 last updated on 02/Jun/24

AC is

$$\mathrm{AC}\:\mathrm{is}\:\:\cancel{\underbrace{\:}} \\ $$

Commented by mr W last updated on 02/Jun/24

yes, since ∠B=90°

$${yes},\:{since}\:\angle{B}=\mathrm{90}° \\ $$

Commented by mr W last updated on 02/Jun/24

Commented by mnjuly1970 last updated on 02/Jun/24

grateful master very nice as always. ⋛

$${grateful}\:\:{master}\: \\ $$$${very}\:{nice}\:{as}\:{always}.\:\underline{\underbrace{\lesseqgtr}} \\ $$

Commented by Tawa11 last updated on 21/Jun/24

Weldone sir

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

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