Question

la ex 4, la b) cum demonstrez ca sunt coliniare ?
si ex 5, va rog!!​

Answer 1

4. b.

• Aplicam Pitagora in ΔABP

BP²=AP²+AB²

BP²=4+12=16

BP=4

• Observam ca 2×AP=BP⇒ m(∡ABP)=30°⇒ m(∡APB)=60°

m(∡MPB)=120+60=180°⇒M,P,B coliniare

5. a.

CD=60m

AB=120m

MB=(120-60):2=30cm⇒ AM=120-30=90cm

• Aplicam Teorema inaltimii in ΔACB dr

CM²=AM×MB

CM²=90×30

CM=30√3m

$A_{ABCD}=\frac{(CD+AB)\times CM}{2} =\frac{(60+120)\times 30\sqrt{3} }{2} =2700\sqrt{3}$

1ha=10 000m²

0,5ha=5000m²

2700√3  <5000

b.

• N simetricul lui B fata de M⇒BM=MN, CM⊥BNΔCNB isoscel

• In ΔCMB aplicam Pitagora

BC²=MB²+CM²

BC²=2700+900=3600

BC=60

2MB=BC⇒ ∡MCB=30°⇒∡CBM=60°⇒ ΔCNB echilateral

NB=NM+MB=30+30=60⇒AN=120-60=60⇒AN=NB

AN=NB

AD=BC

∡A=∡B⇒ ΔDAN≡ΔCBN⇒ ΔDANechilateral⇒∡ADN=60°

• In trapez avem ∡A=∡B=60°⇒ ∡D=120°

∡CDN=∡D-∡ADN=120-60=60°⇒∡ADN=∡CDN⇒DN bisectoarea ∡ADC