Question

problemele din poza, 3b,4 ,5, 6 sau măcar una sau doua daca puteți va rog frumos ​

Answer 1

Răspuns:

Explicație pas cu pas:

a)

3AB= 4BC→AB= 4BC/3

AD= 2BC

5DA= 6 CD →6DA= 5*2BC

DA= (5*2 BC ) /6

²⁾4BC/3+⁶⁾2BC +( 5*2BC) /6+⁶⁾ BC= ⁶⁾360 °

8BC+12BC+ 10BC +6 BC=2160

36BC=2160

BC= 2160:36

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BC= 60 °

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b)

AB= 4BC/3

AB= 4*60/3

AB= 240:3

AB= 80 °

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AD= 2BC

AD= 2*60

AD= 120 °

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6CD=5*120

6CD=600

CD=600:6

CD= 100 °

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∡ABC= (arc CD + arc AD )/2

∡ABC= (100+120 )/2

∡ABC= 220/2

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∡ABC= 110 °

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∡BCD= ( arc AD +arc AB )/2

∡BCD= ( 120+80)/2

∡BCD= 200/2

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∡BCD= 100 °

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∡CDA = ( arc AB + arc BC )/2

∡CDA= (80+60)/2

∡CDA= 140/2

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∡CDA=70°

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∡DAB= ((arc BC +arc CD) /2

∡DAB= (100+60)/2

∡DAB= 160/2

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∡DAB= 80 °

-

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5)

MP║AB→ΔABC asemenea ΔMPC

                   ↓

AB/MP =AC/ MC

AM/MC =1/3→AC/MC= 4/3

atunci

AB/MP =4/3 →20/MP= 4/3

$MP= 20*3/4

MP= 60/4

MP=15 cm

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MQ║BC →ΔAQM asemenea Δ ABC

                     ↓

BC/QM =AC/AM

16/ QM= 4/1

4QM= 16*1

QM= 16/4

QM=4 cm

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Perimetrul MPBQ= 2*( MP+QM) =2*( 15+4) = 2*19= 38 cm

P=38 cm