Răspuns :
20:4=5 latura romb
notam cu o intersectia diagonalelor
se formeaza triunghiul AOB dpretunghic in o
diagonala AC este 6,deci AO este jumate ,adica 3
in triunghiul AOB : AB²=AO²+OB²
5²=3²+AO²
AO²=25-9
AO²=16
AO=4
AC=AO X 2
AC=4X2
AC=8
ARIA = PRODUSUL DIAGONALELOR SUPRA 2
= [tex] \frac{6X8}{2} = \frac{48}{2} =24[/tex]
notam cu o intersectia diagonalelor
se formeaza triunghiul AOB dpretunghic in o
diagonala AC este 6,deci AO este jumate ,adica 3
in triunghiul AOB : AB²=AO²+OB²
5²=3²+AO²
AO²=25-9
AO²=16
AO=4
AC=AO X 2
AC=4X2
AC=8
ARIA = PRODUSUL DIAGONALELOR SUPRA 2
= [tex] \frac{6X8}{2} = \frac{48}{2} =24[/tex]
ABCD-romb
AB=BC=DC=AD=l
BD=6
P=4l
P=20
20=4l
l=20/4=5
A ΔABD=√p(p-a)(p-b)(p-c)
p=(a+b+c)/2= (5+5+6)/2= 16/2=8
A ΔABD=√8(8-5)(8-5)(8-6)= √2³ *3 *3 *2=
A ΔABD= 2² *3= 4*3=12
A ABCD=2* A ΔABD =2 *12=24 cm²
AB=BC=DC=AD=l
BD=6
P=4l
P=20
20=4l
l=20/4=5
A ΔABD=√p(p-a)(p-b)(p-c)
p=(a+b+c)/2= (5+5+6)/2= 16/2=8
A ΔABD=√8(8-5)(8-5)(8-6)= √2³ *3 *3 *2=
A ΔABD= 2² *3= 4*3=12
A ABCD=2* A ΔABD =2 *12=24 cm²