Fie BE_|_CD , E € CD
BE=AD=4cm
DE=AB=9cm
EC=CD-DE=12-9=3cm
În triunghiul CEB, E=90°, aplicăm teorema lui Pitagora
[tex] {bc}^{2} = {ce}^{2} + {be}^{2} [/tex]
[tex] {bc}^{2} = {3}^{2} + {4}^{2} = 9 + 16 = 25[/tex]
BC=5cm
a) perimetrul=BC+AB+AD+CD=5+9+4+12=30 cm
b)
[tex]aria = \frac{(dc + ab) \times h}{2} = \frac{(12 + 9) \times 4}{2} = 21 \times 2 = 42 {cm}^{2} [/tex]
c)
[tex]aria \: abc = \frac{l \times h}{2} = \frac{distanta \times ac} {2} = \frac{ab \times ad}{2} = \frac{9 \times 4}{2} = 9 \times 2 = 18[/tex]
AC×distanta=18×2=36
Aplicăm teorema lui Pitagora in triunghiul ADC
[tex] {ac}^{2} = {ad}^{2} + {dc}^{2} = {4}^{2} + {12}^{2} = 16 + 144 = 160 [/tex]
[tex]ac = 4 \sqrt{10} [/tex]
[tex]distanta = \frac{36}{4 \sqrt{10} } = \frac{4}{ \sqrt{10} } = \frac{4 \sqrt{10} }{10} = \frac{2 \sqrt{10} }{5} [/tex]
