Question

10 si 11 va rog!! ofer coroana la primul care raspunde corect!

Answer 1

ex.10

6=(20-CD) : 2

12=20-CD

CD=20-12=8 cm

(20+8) :2 = 14 cm -semisuma bazelor

ex.11

Eu am notat trapezul ABCD

AB II CD

MN-linia mijlocie

(AB+CD ) :2= 27

AB+CD=27x2=54

(AB-CD)/2=8    ⇒ AB-CD=8x2=16

AB+CD=54

AB=54-CD

54-CD-CD=16

2CD=70

CD=70/2=35

AB=54-35=19

Answer 2

10. linia mijlocie = (AB+DC)/2 = (20+15)/2 = 35/2 = 17,5 cm ;

lungimea segmentului dintre diagonale = (AB-CD)/2 = (20-15)/2 = 5/2 = 2,5 cm

11. (B+b)/2 = 27 ; B+b = 54 ; (B-b)/2 =8 ; B-b = 16 ; adunam cele 2 relatii si obtinem

B+b+B-b = 54+16 ; 2B = 70 ; B = 70/2 = 35 ; b = 54-35 = 19 ;

trapezul are baza mare = 35 cm si baza mica = b = 19 cm ;