Question

Intro piramida patrulatera regulata inaltimea are lungimea de3 cm, iar muchia laterala de 5 cm sa se afle volumul piramidei

Answer 1

VO⊥(ABC)
OC ⊂ (ABC)   ⇒ VO ⊥ OC 
                       
⇒ΔVOC , m(o) = 90°
⇒OC² = 5² - 3² 
   OC² =25 - 9
   OC = 4 cm

AC = 4·2 = 8 cm
ΔABC dr isoscel
AC² = x² +x²
64 = 2x² 
x² = 32
x = 4√2 cm  (X = AB = BC )

Aria bazei = (4√2)² 
                  =32cm²

V = Ab ·h /3
V = 32 ·3 /3 
V = 32 cm³

Answer 2

semidiagonala bazei=(Pitagora) =√(5²-3²)=4cm⇒diagonal bazei=8cm⇒aria bazei=8*8/2 (patratul e si romb)=32cm²
V=(1/3) *32*3=32cm³
as simple as that!!