Cu teorema lui Pitagora în ΔADC ⇒ DC = 12,8cm
BD = BC - DC = 20 - 12,8 = 7,2cm
Fie DE = d(D, AC), DF = d(D, AB) ⇒ DE, DF sunt înălțimi în
ΔADC, respectiv ΔABD
[tex]\it \mathcal{A}_{ADC}=\dfrac{c_1\cdot c_2}{2}=\dfrac{AD\cdot DC}{2}= \dfrac{9,6 \cdot12,8}{2}\\ \\ \\ \mathcal{A}_{ADC} =\dfrac{AC\cdot DE}{2} \Rightarrow \dfrac{9.6\cdot12,8}{2}=\dfrac{16\cdot DE}{2} \Rightarrow DE=\dfrac{9,6\cdot12,8}{16}=7,68cm[/tex]
[tex]\it \mathcal{A}_{ABD}=\dfrac{c_1\cdot c_2}{2}=\dfrac{AD\cdot BD}{2}= \dfrac{9,6 \cdot7,2}{2}\\ \\ \\ \mathcal{A}_{ABD} =\dfrac{AB\cdot DF}{2} \Rightarrow \dfrac{9.6\cdot7,2}{2}=\dfrac{12\cdot DE}{2} \Rightarrow DE=\dfrac{9,6\cdot7,2}{12}=5,76cm[/tex]
DF + DE = 5,76 + 7,68 = 13,44 cm
