BC=21 cm
M∈BD (avand in vedere ca nu ai specificat, eu asa presupun)
BD=AC (diagonale dreptunghi)
[tex]sin=\frac{cateta \ opusa}{ipotenuza}[/tex]
[tex]sin MBC=sin DBC=\frac{DC}{BD}=\frac{3}{5}[/tex]
5DC=3BD
Aplicam Pitagora (suma catetelor la patrat este egala cu ipotenuza la patrat)
BD²=DC²+BC²
[tex]\frac{25DC^2}{9}=DC^2+ 441[/tex]
16DC²=9×441
[tex]DC=\frac{63}{4}[/tex]
[tex]5\times \frac{63}{4}=3BD \\\\BD=\frac{105}{4}=AC[/tex]
b)
[tex]A=BC\times DC=21\times \frac{63}{4} =\frac{1323}{4}\ cm^2[/tex]
c)
[tex]A_{BDC}=A_ABCD}:2=\frac{1323}{8}[/tex]
[tex]A_{BDC}=A_{BMC}+A_{DMC}\\\\Dar\ A_{DMC}=A_{BMA}[/tex]
[tex]A_{BMC}=\frac{h\times BC}{2} =\frac{\frac{63}{8} \times 21}{2} =\frac{1323}{16}[/tex]
[tex]A_{BMA}=\frac{1323}{8}-\frac{1323}{16}=\frac{1323}{16}[/tex]
Deci raportul va fi egal cu 1
[tex]\frac{A_{BMC}}{A_{ABM}} =1[/tex]
