Question

exercitiul 5,6 va rogg !

Answer 1

6) duci CM║DA⇒MB=√15²-9²⇒√144=12
AB=12+8+20
Ar=(B+b)*h/2⇒(20+8)*9/2⇒14*9=126

5) diagonalele se intretaie in parti egale
aflam a doua diagonala (pitagora) 26²=24²+AO²
AO=√ 26²-24²=√10=10

A=D*d/2=48*10/2=240cm²

Answer 2

5.
BD=48cm
$BO= \frac{BD}{2}=24$
$In$ Δ $BOC(teorema lui Pitagora)=\ \textgreater$
$BC^{2} = BO^{2} +CO^{2}$
$CO ^{2} = BC^{2} - BO^{2}$
$CO^{2} =676-576$
$CO^{2} =100$
$CO=10cm$

$AC=CO*2=20cm$

$A_{ABCD} = \frac{ d_{1}* d_{2} }{2} = \frac{BD*AC}{2} = \frac{48*20}{2} =480 cm^{2}$
6.
$DC=8cm$
$DA=CM=9cm$
Δ$CMB(dreptunghic)=CB=15cm$
$In$ Δ $BMC(teorema lui Pitagora)=\ \textgreater$
$CB^{2} = CM^{2} + MB^{2}$
$MB^{2} = CB^{2} - CM^{2}$
$MB^{2} = 225-81$
$MB^{2} =144$
$MB=12cm$

$AB=AM+MB$
$AB=8+12$
$AB=20cm$

$A_{ABCD} = \frac{(b+B)*h}{2}= \frac{(AB+DC)*CM}{2} = \frac{28*9}{2} =14*9=126cm^{2}$

O zi buna! Sper ca ti-am fost de ajutor! :)