In trapezul ABCD , AB paralel CD , AB = 21 cm , BC = 15 cm , CD = 7 CM si AD = 13 
CALCULATI:
a) inaltimea 
b) lungimile diagonalelor trapezului

Răspuns :

a) Notam DM_|_AM si CN_|_AB
=> DM si CN sunt inaltimi
h²=DM²=AM²-13²
h²=CN²=BN²-15²
=>AM²-13²=BN²-15²
dar
AM+BN=21-7=14
AM=14-BN
AM²=(14-BN)²=14²+BN²-2*14*BN
=>14²+BN²-2*14*BN-13²=BN²-15²
196+BN²-28BN=BN²-225-169
28BN=252 =>BN=9

h²=CN²=BC²-BN²=15²-9²=225-91=144
h=12

b)In ΔMDB , avem
MB=7+9=16
DB²=MD²+MB²=12²+16²=144+256=400
DB=20

In ΔANC
AN=21-9=12
AC²=AN²+NC²=12²+12²=144+144=228
AC=12√2
Vezi imaginea Cpw