testul B problema 1
ducem AD_I_BC
triunghiul ABD dreptunghic isoscel ⇒ BD=AD=3
triunghiul ADC dreptunghic
DC=BC-BD=4-3=1
AC=√(3²+1²)=√10
P=3√2+4+√10
3.
notam ipotenuza cu a
notam catetele cu c si b
Atriunghi =bxc/2
Asemicercurilor
π(a/2)²/2=a²π/8
π(b/2)²/2=b²π/8
π(c/2)²/2=c²π/8
Afigurii hasurate =bxc/2+b²π/8+c²π/8-a²π/8=bxc/2+π/8(b²+c²-a²)
dar a²=b²+c²
deci
bxc/2+π/8(b²+c²-a²)=bxc/2(adica aria triunghiului)
test C pr.1.
ducem AD_I_BC
triunghiul ABD dreptunghic isoscel
AD=BD=√16/2=√8=2√2
sin 30⁰=AD/AC
1/2=2√2/AC
AC=2x2√2=4√2
cos 30⁰=DC/AC
√3/2=DC/4√2
DC=√3x4√2/2=2√6
BC=2√2+2√6=2√2(1+√3)
2) m(<A)=3k
m(<B)=6k
3k+6k=90⁰
k=10
m(<A)=3k=30⁰
m(<B)=6k=60⁰
sin 30⁰=CB/AB
1/2=CB/36
CB=18 cm
AC=√(36²-18²)=√(1296-324)=√972=18√3 cm
3.
din calculul elementelor in poligoane regulate avem formule in functie de raza
a)
L₃=R√3=12√3 cm
a₃=R/2=12/2=6 cm
A₃=3R²√3/4=3x12²√3/4=108√3 cm²
b)
L₄=R√2=12√2 cm
a₄=R√2/2=12√2/2=6√2 cm
A₄=2R²=2x12²=288 cm²
P₄=4x12√2=48√2 cm
