Pyramid - angle

Calculate the surface of regular quadrangular pyramid whose base edge measured 6 cm and the deviation from the plane of the side wall plane of the base is 50 degrees.

Result

S =  92.006 cm2

Solution:

h=(a/2)tan50=3.575 cm h2=h2+(a/2)2=4.667 cm S1=ah2/2=14.001 S=4S1+a2=92.006 cm2 h = (a/2) \cdot \tan 50 ^\circ = 3.575 \ cm \ \\ h_2 = \sqrt{ h^2 + (a/2)^2 } = 4.667 \ cm \ \\ S_1 = a h_2 / 2 = 14.001 \ \\ S = 4 \cdot S_1 + a^2 = 92.006 \ \text{cm}^2



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