# 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:

$h = (a/2) \cdot \tan 50 ^\circ = 3.575 \ cm \ \\ h_2 = \sqrt{ h^2 + (a/2)^2 } = 4.667 \ cm \ \\ S_1 = a h_2 / 2 = 14.001 \ \\ S = 4 \cdot S_1 + a^2 = 92.006 \ cm^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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