Question

Trapezul dreptunghic ABCD are m(A)=m(D)=90°, AB>CD , AC perpendicular pe BC, iar MN linie mijlocie (M apartine AD, N apartine BC) ,MN int AC={P} si MN int BD ={Q} . Stiimd ca AD =18 cm si PQ = 12 cm , calculeaza P ABCD =? , lungimile diagonalelor AC si BD =? , distanta de la punctul D la dreapta BC =? ajutorrr dau coroana imd

Answer 1

AD=18
 PQ=(AB-CD)/2=12 =>AB-CD=24
Se formeaza un  ?dreptunghic ducand o perpendiculara din C pe AB, CF in care:
FB=24, CF=18
=> CB²=FB²+CF²=24²+18²=576+324=900
CB=30

Cf informatii suplimentare AC_|_BC=>
In  Δdreptunghic ACB, cf teoremei catetei, :
CB²=FB*AB
30²=24*AB²
900=24*AB
AB=900/24
AB=75/2=37.5
CD=AB-24=72/2-24=72/2-48/2
CD=27/2=13.5

Perimetru Trapez ABCD=AB+BC+CD+DA=72/2+30+27/2+18=99

Lungimile diagonalelor AC si BD
AF=DC
AC²=AF*AB=27/2*75/2=2025/4=506.25
AC=45/2=22.5

BD²=AD²+AB²=18²+(75/2)²=324+5625/4=1296/4+5625/4=6921/4=769*9/4
BD=3√769/4


 Distanta de la d la dreapta bc
fie DE_|_BC
Aria ΔCDB=DC*DA/2=CB*CE/2
=> 27/2*18=30*CE
27*9/30=CE
CE=81/5

sper ca e bine