[tex]\bf AD \perp BD \implies \angle ADB =90^{\circ}[/tex]
[tex]\bf In \;\; \triangle ABD \; \; avem:[/tex]
[tex]\bf AD = 6\, cm \; (cateta)[/tex]
[tex]\bf AB = 10\, cm \; (ipotenuza)[/tex]
[tex]\bf BD = 8 \, cm \; (cateta)[/tex]
[tex]\bf \implies h= \dfrac{c_1 \cdot c_2}{ip} =\dfrac{6 \cdot 8 }{10 } = 4,8 \, cm[/tex]
[tex]\bf A_{ABCD} = b \cdot h = 10 \, cm \cdot 4,8 \, cm[/tex]
[tex]\bf \implies \boxed{\bf A_{ABCD}=48\, cm^2}[/tex]
#copaceibrainly
