Răspunsuri

2014-08-02T15:51:09+03:00
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
 \frac{6X8}{2} = \frac{48}{2} =24

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2014-08-02T15:51:59+03:00
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


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