Question

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Answer 1

4.∆ABC dreptunghic in A

AB=45cmAC=60cm

TPitagora BC=√AB²+AC²=√45²+60²=75cm

M€AC a î.AM/CM=1/5 =>

CM=5AM =>

a)AC=60cm=CM+AM=5AM+AM=6AM

AM=60/6=10cm

MN_l_BC

b) perimetrul ∆MNC=CM+MN+NC

CM=AC-AM=60-10=50cm

∆MNC≈∆ABC {<C comun;dreptunghice}

=>rapoartele de asemănare

BC/MC=AB/MN=AC/NC==>

75/50=45/MN=60/NC

3/2=45/MN

MN=45×2/3=30cm

3/2=60/NC

NC=60×2/3=40cm

perimetrul ∆MNC =30+40+50=120cm

5. ∆ABC <B=60°. BD bisectoarea lui

AB=8√3cm. BD =8√3cm

a)<BDC=?

∆ABDisoscel AB=BD

=><A=<D. <B/2=30°

<A=(180-30)/2=150/2=75°

<BDC este exterior=

<B/2+<A=30°+75°=105°

b) Aria ∆BDC

observăm în ∆ABC

<C=180-60-75=45°

B/2=30°

teorema sinusurilor

DC/sin30°=BD/sin45

BD=8√3

DC/(1/2)=8√3/(√2/2)

DC=8√3/√2=8√6/2=4√6cm

aria=DC×BDsin105°/2=

(4√6×8√3)/2sin105=

6×8√2×(√6+√2)/4=12√2√6+12√2√2=

24√3+24=24(√3+1)cm²