Question

In trapezul isoscel MNPQ cu NP||MQ se stie ca :MN=6,5 ; NP=4cm si MQ=9 cm.Calculează:a)Aria trapezului b)lungimea diagonalei

Answer 1

MN = PQ = 6 . 5 CM

( baza mica + baza mare ) / 2 = ( 4 + 9 ) / 2 = 6.5 cm

$h {}^{2} = (6.5cm {}^{2} ) {}^{2}$
=
$= 42.25cm {}^{2} - 6.25cm {}^{2}$
$= 36cm {}^{2}$
$h = \sqrt{} 36cm {}^{2} = 6cm$
a ) Aria trapezului = linia mijlocie X inaltime = 6.5 cm × 6 cm =
$= 39cm {}^{2}$
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
$mp { = }^{2} = ps {}^{2} + ms {}^{2} =$
= 36 + 42 , 25 = 78 , 25 => MP =
$5 \sqrt{313}$