Question

Geometrie, a VIII- a, teorema celor 3 perpendiculare.


Pe planul triunghiului dreptunghic ABC, m(<A)=90*, cu AB=30 cm, AC=40 cm se ridica perpendiculara AP, cu AP=8√3 cm. Aflati:
a) distanta de la P la BC
b) distanta de la A la planul PBC
c) masura unghiului diedru format de planele PBC si ABC.

Answer 1

[tex]AB^2 + AC^2 = BC^2 => BC = \sqrt{AB^2 + AC^2} = \sqrt{40^2+30^2} \\ =\sqrt{2500}=50 \\\\a)\: AC*AB = BC*AM => AM = \frac{AC*AB}{BC} = 32 \\\\ \:Distanta\:de\:la\:P\:la\:BC\:este\:tocmai\:perpendiculara\:PM\:unde\:M\in BC\\ \:In\:triunghiul\:APM:\\ PA \perp AM => \angle PAM = 90^\circ => PAM\:triunghi\:dreptunghic\\ => PM^2 = AP^2 + AM^2 => PM = \sqrt{AP^2 + AM^2} = \sqrt{64*3 + 32^2} \\ = \sqrt{1216} = 4\sqrt{71} \\\\ \:Distanta\:de\:la\:A\:la\:planul\:PBC\:este\:tocmai\:perpendiculara\:AN[/tex]
[tex]sa\:spunem, unde\:N \in PBC\:si\:chiar\:mai\:mult, N\:\in PM\\\\ Aplicam\:ca\:si\:mai\:sus\:formula\:ariei\:care\:se\:poate\:scrie\:in\:2\:moduri\\ pentru\:un\:triunghi\:dreptunghic,\:adica\:baza\:ori\:inaltimea\:supra\:2\\ sau\:cateta\:ori\:cateta\:supra\:2.\\ AM*AP = AN*PM => AN = AM*AP / PM = \frac{32 * 8 \sqrt{3}}{4 \sqrt{71}} = \frac{64 \sqrt{213} }{71} \\\\[/tex]