Question

In trapezul ABCD AB=5 cm CD=20cm AD =9cm BC 12 cm a?)aratati ca inaltimea trapezuli are 7,2 cm b) aflati aria trapezului ABCD c) aflati distanta De la D la BC

Answer 1

a)
ducem AM_I_DC    DM=x
ducem BN_I_DC    NC=y
AM=BN=h
h²=9²-x²
h²=12²-y²
9²-x² =12²-y²
y²-x² =12²-9²
(y-x)(y+x)=(12-9)(12+9)
(y-x)(y+x)=63
x+y=20-5=15
y-x=63/15=4,2
2y=15+4,2=19,2
y=19,2/2=9,6 cm
x=15-9,6=5,4 cm       
h=√(9²-5,4²)=√(81-29,16)=√51,84=7,2 cm

b)
Atrapez=(20+5)×7,2/2=25×3,6=90 cm²

c)
Atriunghi DBC=20×7,2/2=72 cm²
Atriunghi DBC=12×d(D,BC)/2=72 cm²
d(D,BC)=72×2/12=12 cm