Question

Ex.1,2,3
Test 1
P.S Dau 54 puncte

Answer 1

1)NP²=MN²+MP=18²+24²=324+576=900
NP=√900=30
MN²=QN·PN
18²=30·QN
324=30·QN
QN=324/30=10,8
PQ=30-10,8=19,2
MQ²=QN·PQ=10,8·19,2
MQ=14,4

2)AD²=AB²-BD²=20²-16²=400-256=144
AD=12
A=AD·BC/2=12·32/2=192
A=AC·BE/2
192=20·BE/2
192=10BE
BE=19,2

3) triunghiul ACB dreptunghic
fie CM inaltimea triungiului
MB²=(4√6)²-(4√2)²=96-32=64
MB=8
aplicam teorema inaltimii
CM²=AM·MB
32=AM·8
AM=4
AB=AM+BM=4+8=12 baza mare
AM=DC=4 baza mica
AC²=(4√2)²+4²=32+16=48
AC=√48=4√3
BD²=(4√2)²+12²=32+144=176
BD=√176=4√11
DE⊥AC
DE=DC·DA/AC=4·4√2/4√3=4√6/3