Question

Intr-un trapez isoscel ABCD (ABIICD),AC perpendicular pe BC,AC=8 cm si BC=4cm.

a) realizati un desen corespunzator
b) Calculati perimetrul trapezului
c) Calculati inaltimea trapezului
d) Aflati distanta de la punctul C la latura AD.

Answer 1

din ipoteza AC_|_BC⇒ΔACB-dreptunghic

ΔACB,mas<C=90  ⇒AB²=AC²+BC²=8²+4²=64+36=100  ⇒AB=10
fie CN_|_AB
CN-h    ⇒
⇒CN=AC*BC/AB=  8*4/10= 4*4/5= 16/5

ΔNBC,mas<N=90  ⇒NB²=BC²-CN²= 4²- (16/5)²= 16-  256/25   aducem la acelasi numitor
NB²=(400-256)/25= 144/25   ⇒NB=12/5

fie DM_|_AB
DM-h    ⇒DC=MN=AB-2NB= 10- 2*12/5= 10- 24/5  aducem la acelasi numitor

DC=MN=(50-24)/5= 26/5

P=AB+DC+2BC= 10+ 26/5 + 2*4  aducem la acelasi numitor
P=(50+26+40)/5
P=116/5

Answer 2

raspunsul in fisierul atasat