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2014-04-12T18:59:51+03:00
Duci CN_/_AB  DC=AN=24
NB=AB-AN=36-24=12
ΔBNC, mas<N=90  cos <B=60=√1/2=1/2
cos<B=NB/BC
12/BC=1/2
BC=2*12=24
CN²=BC²-NB²=24²-12²=576-144=432  ⇒CN=12√3
CN=AD=12√3

P=24+24+36+12√3=84+12√3
A=(B+b)*h/2= (24+36)*12√3/2=  60*6√3=360√3
DB-diag ΔDCB-isoscel ⇒DB=24√2
AC-diag
ΔANC, mas<N=90  AC²=AN²+CN²=24²+(12√3)²=576+432=1008  ⇒AC=12√7

2014-04-12T20:34:03+03:00
ABCD-TRAPEZ DREPTUNGHIC in A si D
AB=36 cm,DC=24 cm, m(unghi ABC)=60 GRADE
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AC=?   DB=?  P=? S=?
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ADparalel cu CM=12 RADICAL DIN 3,
TRIUNG     MBC    MB=AB-DC=36-24=12cm , m(ung CMB)=90, m(ung MBC)=60=> m( ung BCM)=30
SI cf t latura care se opune ung de 30 grade intr-un tr dreptung este 1/2 din ipot, cum MB=12 cm=> BC=24 cm, din T lui PITAGORA MC=12 RADICAL 3 cm ,
din tr     DAB dreptunghic in A  AB=36 cm, AD=12 radical 3 cm, DB =24 RADIC 3cm( din T lui PITAGORA)
Din tr ACM DREPTUNG IN M, MC=12 radical 3 cm, AM= 24 cm, =>AC=12 RADICAL 7cm ,
P=AB+BC+DC+AD=36+24+24+12RADICAL 3=(84+12radical 3 )cm
S=(AB+DC)*AD/2=360radical 3 cm patrati, unde P-perimetrul trapez, S- aria trap, AC, DB- diagonale