Răspuns :
Notam ABCD trapezul
AB II CD
linia mijlocie eate MN
(AB+CD ) :2= 27 ⇒ AB+CD=27x2=54
segmentul care uneste mijloacele diagonalelor este 8 este PQ
(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
linia mijlocie e (b+B) impartit la 2 = 35÷2=17,5
segmentul are si el o formula adica semidiferenta bazelor : (20-15)÷2 = 2,5
AB II CD
linia mijlocie eate MN
(AB+CD ) :2= 27 ⇒ AB+CD=27x2=54
segmentul care uneste mijloacele diagonalelor este 8 este PQ
(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
linia mijlocie e (b+B) impartit la 2 = 35÷2=17,5
segmentul are si el o formula adica semidiferenta bazelor : (20-15)÷2 = 2,5
ABCD trapez AB II CD MN-lin mij
(AB+CD ) :2= 2
AB+CD=27x2=54
FG=8
(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
lin mij=b+B: 2 = 35:2=17,5⇒ (20-15):2 = 2,5
(AB+CD ) :2= 2
AB+CD=27x2=54
FG=8
(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
lin mij=b+B: 2 = 35:2=17,5⇒ (20-15):2 = 2,5