Xét ΔABC có:
\(BC^2=AB^2+AC^2 \\ =5,4^2+7,2^2\\ =81\\ \Rightarrow BC=\sqrt{81}=9\left(cm\right)\)
\(MB=\dfrac{1}{2}BC\left(gt\right)\\ =\dfrac{1}{2}9\\ =4,5\left(cm\right)\)
Xét ΔABC và ΔEMB có:
\(\widehat{EMB}=\widehat{CAB}=90^0\\ \widehat{B}chung\)
→ ΔABC ∼ ΔEMB(g-g)
\(\rightarrow\dfrac{AB}{BM}=\dfrac{AC}{EM}=\dfrac{BC}{EB}hay\dfrac{5,4}{4,5}=\dfrac{7,2}{EM}=\dfrac{9}{EB}\\ \Rightarrow EM=\dfrac{7,2.4,5}{5,4}=6\left(cm\right)\\ \Rightarrow EB=\dfrac{9.4,5}{5,4}=7,5\left(cm\right)\\ \Rightarrow EB+EM=6+7,5=13,5\left(cm\right)\)
☛Chọn C. 13,5 cm
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