Răspuns :
a)construiesti CN[tex] \perp[/tex] AB⇒CN=AD=160
BN=AB-CD=120m
ΔBNC:m(∧BNC)=90°⇒BC²=BN²+CN²⇒BC=√40000=200m
Lgard=[tex] P_{ABCD} [/tex]=AB+BC+CD+AD=560m
b) ΔAMB:m(BAM)=90°⇒ΔAMB-dreptunghic
ΔCDM:m(CDM)=90°⇒ΔCDM-dreptunghic
pentru ΔBMC calculezi cu Teorema lui Pitagora:
in ΔABM pe BM si va fi BM=√32000=80√5
in ΔCDM pe CM si va fi CM=√8000=40√5
BC²=40000
BM²=32000
CM²=8000
BC²=CM²+BM²⇒ΔBMC-dreptunghic in M (reciproca Teoremei lui Pitagora)
c)[tex] A_{ABM}= \frac{AB*AM}{2}= \frac{160*80}{2}=6400 m^{2} [/tex]
[tex] A_{CMD}= \frac{CD*MD}{2}= \frac{40*80}{2}=1600 [/tex]
[tex] A_{BMC}= \frac{CM*BM}{2}= \frac{40 \sqrt{5}*80 \sqrt{5} }{2}=8000 m^{2} [/tex]
BN=AB-CD=120m
ΔBNC:m(∧BNC)=90°⇒BC²=BN²+CN²⇒BC=√40000=200m
Lgard=[tex] P_{ABCD} [/tex]=AB+BC+CD+AD=560m
b) ΔAMB:m(BAM)=90°⇒ΔAMB-dreptunghic
ΔCDM:m(CDM)=90°⇒ΔCDM-dreptunghic
pentru ΔBMC calculezi cu Teorema lui Pitagora:
in ΔABM pe BM si va fi BM=√32000=80√5
in ΔCDM pe CM si va fi CM=√8000=40√5
BC²=40000
BM²=32000
CM²=8000
BC²=CM²+BM²⇒ΔBMC-dreptunghic in M (reciproca Teoremei lui Pitagora)
c)[tex] A_{ABM}= \frac{AB*AM}{2}= \frac{160*80}{2}=6400 m^{2} [/tex]
[tex] A_{CMD}= \frac{CD*MD}{2}= \frac{40*80}{2}=1600 [/tex]
[tex] A_{BMC}= \frac{CM*BM}{2}= \frac{40 \sqrt{5}*80 \sqrt{5} }{2}=8000 m^{2} [/tex]
