Question

Exercitiile sase si sapte

Answer 1

6.
ducem CF perpendicular pe AB
CF=AD=12 cm
FB=AB-CD=16 cm
teorema inaltimii
CF=√AF×FB
12=√AF×16
144=AF×16
AF=144:16=9 cm
AF=DE=9 cm
AB=9+16=25 cm
Atrapez=(25+9)×12/2=204 cm²

7.
E=√(2x-3)²+√(x-4)²
E(-√3)=√[2×(-√3)-3]²+√[(-√3)-4]²=√(-2√3-3)²+√(-√3-4)²= =2√3+3+√3+4=7+3√3
E(√3)=√(2√3-3)²+√(√3-4)²=2√3-3+4-√3= 1+√3
a=E(-√3)-3E(√3)=7+3√3-3( 1+√3)=7+3√3-3-3√3=4