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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

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

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X, Y and Z are points on the sides AB, BC and AC of the triangle ABC, such that AX:XB =4:3, BY:YC=2:3, CZ:ZA=2:1. Find the ratio of the area of the triangle XYZ to that of triangle ABC.

$${X},\:{Y}\:{and}\:{Z}\:{are}\:{points}\:{on}\:{the}\:{sides}\:{AB}, \\ $$$${BC}\:{and}\:{AC}\:{of}\:{the}\:{triangle}\:{ABC},\:{such} \\ $$$${that}\:{AX}:{XB}\:=\mathrm{4}:\mathrm{3},\:{BY}:{YC}=\mathrm{2}:\mathrm{3},\: \\ $$$${CZ}:{ZA}=\mathrm{2}:\mathrm{1}.\:{Find}\:{the}\:{ratio}\:{of}\:{the}\:{area} \\ $$$${of}\:{the}\:{triangle}\:{XYZ}\:{to}\:{that}\:{of}\:{triangle} \\ $$$${ABC}. \\ $$

Question Number 207691    Answers: 3   Comments: 0

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