Question

1. Fie trapezul ABCD AB | | CD, AB = 9 cm, CD = 4 cm, AD= 3 cm si BC = 4 cm. Dacǎ
DE | | BC, E AB :
a) Demonstraţi cǎ triunghiul ADE este triunghi dreptunghic
b) Aflaţi aria trapezului ABCD
c) Aria lui DEBC

Answer 1

a)Avem,DEllBC
Mai,avem,ABllCD
=>DE=BC=4cm
AD=3cm(ip.)
Daca,ABllCD=>EBllCD=>EB=CD=4am
=>AE=9-EB=5cm
Trebuie,sa,vedem,daca,t.Pit.in,ΔADE,este,realizabila
AD²+DE²=?AE²
3²+4²=?5²
9+16=25(A)=>ΔADEdreptunghic

b)Coboram,inaltimile,CC',si,DD'
Stim,ca,DD'=DC=4cm
Notam,segmentele,AD'=x,si,C'B=y
x+y=9-4
x+y=5
Din,Δdr.AD'D,si,Δdr.CC'B,calculam,inaltimile,h,care,sunt,egale.
Avem,h²+x²=9,si,h²+y²=16
Egalam,cele,doua,expresii,si,avem
9-x²=16-y²
9-x²-16+y²=0
y²-x²=7
(y-x)(y+x)=7
Stim,x+y=5
=>y-x=7/5
Aceste,doua,ecuatii,devin,un,sistem,pe,care,il,putem,rezolva,fie,prin,metoda,reducerii,fie,prin,substitutie.Am,sa,aleg,substitutia(nefiind,sigura,daca,ai,invatat,alta,metoda...)
Din,prima,x=5-y
O,introducem,in,a,doua,y-(5-y)=7/5=>y-5+y=7/5=>2y=7/5+5=>2y=32/5
=>y=16/5
=>x=5-16/5=9/5
Avand,x,si,y,putem,calcula,inaltimea,h,in,trapez
Stim,cu,t.Pit.
h²+(16/5)²=16
h²=16-256/25
h²=(400-256)/25
h²=144/25
=>h=12/5

Aria=(B+b)xh/2=13x12/10=156/10=15,6cm²

c)DEBC,este,paralelogram
A=bazaxh
h-inaltime,comuna,cu,a,trapezului
A=12/5x4=48/5=9,8cm²