MN = PQ = 6 . 5 CM
( baza mica + baza mare ) / 2 = ( 4 + 9 ) / 2 = 6.5 cm
[tex]h {}^{2} = (6.5cm {}^{2} ) {}^{2} [/tex]
=
[tex] = 42.25cm {}^{2} - 6.25cm {}^{2} [/tex]
[tex] = 36cm {}^{2} [/tex]
[tex]h = \sqrt{} 36cm {}^{2} = 6cm[/tex]
a ) Aria trapezului = linia mijlocie X inaltime = 6.5 cm × 6 cm =
[tex] = 39cm {}^{2} [/tex]
EF = ( MN + PQ ) / 2
EF = ( L + 2L ) / 2
EF = 3L / 2
l = 2 EF / 3 = 2 × 6 / 3 = 4 CM
DECI PQ = 8 CM
ATUNCI PERIMETRUL VA FI
3 * L + 2L = 5L = 5 × 4 = 20 CM
B ) DIAGONALA MP FIE DREPTUNCHI PSM => CONFORM TH PITAGORA
[tex]mp { = }^{2} = ps {}^{2} + ms {}^{2} = [/tex]
= 36 + 42 , 25 = 78 , 25 => MP =
[tex]5 \sqrt{313} [/tex]
