MR-692 / 12. zadatak

Osnovice trapeza su 142 cm i 89 cm, a dijagonale su 120 cm i 153 cm. Odrediti površinu trapeza.

Pitagorina teorema za trougao AFC:

h2+(89+x)2=1202

h2=1202-(89+x)2

Pitagorina teorema za trougao EBD:

h2+(142-x)2=1532

h2=1532-(142-x)2

1202-(89+x)2=1532-142-x2

14400-7921+178x+x2=23409-20164-284x+x2

14400-7921-178x-x2=23409-20164+284x-x2

-178x-284x=23409-20164-14400+7921

-462x=-3234 |·-1

462x=3234 

x=3234 462

x=7

x+89+y=142

y=142-x-89

y=142-7-89

y=46

h2=1202-(89+x)2=1202-89+72=1202-962

h2=14400-6724=5184

h=5184

h=72

P=PAED+PEFCD+PFBC

PAED=x·h2

PAED=7·722

PAED=7·722=7·2·362=7·36

PAED=252

PEFCD=b·h=89·72

PEFCD=6408

PFBC=y·h2

PFBC=46·722=2·23·722=23·72

PFBC=1656

P=PAED+PEFCD+PFBC=252+6408+1656

P=8316