Explicație pas cu pas:
notăm DM ⊥ BC, M ∈ BC
ducem AN ⊥ BC, N ∈ BC => AN mediană
BN = ½×BC => BN = 12 cm
T.P.: AN² = AB² - BN² = 36² - 12² = 1152
=> AN = 24√2 cm
BC = BD = 24 cm => ΔBCD isoscel
∢BCD ≡ ∢BDC => ∢BDC ≡ ∢ABC
=> ΔABC ~ ΔBCD
[tex]\frac{AB}{BC} = \frac{36}{24} = \frac{3}{2} \\ [/tex]
[tex]\frac{Aria_{\triangle ABC}}{Aria_{\triangle BCD}} = {\Big( \frac{3}{2} \Big)}^{2} \iff \frac{ \frac{AN \cdot BC}{2} }{ \frac{DM \cdot BC}{2} } = \frac{9}{4} \\ \frac{AN}{DM} = \frac{9}{4} \iff DM = \frac{4 \times 24 \sqrt{2}}{9} \\ \bf \implies DM = \frac{32 \sqrt{2} }{3} \: {cm}^{2} [/tex]
