Question

Se considera piramida triunghiulara regulata VABC cu varful in V care are nuchia bazei de 4 radical 3 cm si aria laterala egala cu 24 radical 3 cm patrati . AFLATI : a) V ; b) masura unghiului dintre planele ( VBC ) SI (ABC) ; c) distanta de la punctul A la planul ( VBC) .

Answer 1

a) $Ab= \frac{ l^{2} \sqrt{3} }{4} = \frac{ (4\sqrt{3})^{2} \sqrt{3} }{4}= \frac{ 48 \sqrt{3} }{4}= 12 \sqrt{3} cm^{2}$

$V= \frac {Ab*h}{3}$

$m^{2} = h^{3} + R^{2} => h^{2} = m^{2} - R^{2} <=> h^{2} = (4 \sqrt{3})^{2} + 4^{2} =$
$= 48 - 16 = 32 => h = \sqrt{32} = 2 \sqrt{2} cm$

$V= \frac{12 \sqrt{3} * 2 \sqrt{2} }{3} = 8 \sqrt{6} cm^{3}$

$Al= \frac{ P_{b}* a_{p} }{2} <=> 24 \sqrt{3} = \frac{12 \sqrt{3}* a_{p} }{2} <=> 24 \sqrt{3} =6 \sqrt{3} * a_{p}  =>$
$=> a_{p} = \frac{24 \sqrt{3} }{12 \sqrt{3} } } =2 cm$

b) m (unghiului (VBC), (ABC)) = m( (VM), (AM) ) = m(OMV)
M∈ (BC)
VM perpendiculara pe BC, VM∈ (VBC)
OM perpendiculara pe BC, OM∈ (ABC)
$sin OMV = \frac{VO}{VM} = \frac{2 \sqrt{2}}{2} = \sqrt{2}$ 

c) d( A, (VAC)) = $pr_{(VBC)} A= M$ = [AM]

In ΔABM ( unghilul M = 90°) ⇒(conform teoremei lui Pitagora) AM² = AB² - BM² ⇔ AM² = (4√3)² - (2√3)² = 48 - 12 = 36 ⇒ AM = √36 = 6 cm

(Scuze de desen; e cam uratel)