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Flura [38]
3 years ago
8

Ragazzi vi presento questo problema è mi piacerebbe se riusciate ad aiutarmi a risolverlo:"Calcola il perimetro di un trapezio i

soscele sapendo che ha l'area di 660 m(quadrati), l'altezza di 12 m e la base minore di 20 m.​

Mathematics
2 answers:
Drupady [299]3 years ago
7 0

Facendo riferimento alla figura allegata, partiamo dall'area: sappiamo che

A=\dfrac{(B+b)h}{2}

dove B e b sono rispettivamente la base maggiore e la base minore, mentre h è l'altezza.

Sostituendo i valori nella formula, possiamo ricavare la base maggiore:

660=\dfrac{12(B+20)}{2}=6(B+20)

Dividendo entrambi i membri per 6 abbiamo

110=B+20 \iff B=90

A questo punto abbiamo

B-b=2AH \iff 2AH=70 \iff AH=35

Conoscendo sia AH sia l'altezza BH, possiamo usare il teorema di Pitagora per calcolare il lato obliquo AB:

AB=\sqrt{AH^2+BH^2}=\sqrt{1225+144}=37

Ricapitolando, abbiamo:

  • AB=37
  • BC=20
  • CD=37 (per simmetria, è uguale ad AB perché il trapezio è isoscele)
  • AD=90

La somma dei quattro lati di darà il perimetro. Ciao!

iren [92.7K]3 years ago
3 0

Answer:

per il vertice basta che usi la formula: V (-b/2a ; - delta/4a)

asse x : -b/2a

Step-by-step explanation:

here is your answer to your work plz rate me the most brainlest

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