The height of a regular quadrangular pyramid forms an angle of 60 ° with the plane of the side face.

| February 4, 2021 | education

The height of a regular quadrangular pyramid forms an angle of 60 ° with the plane of the side face. The lateral surface area of the pyramid is 32√3. Find the volume of the pyramid.

Since the side faces of a regular pyramid are equal triangles, then Scdk = Sside / 4 = 32 * √3 / 4 = 8 * √3 cm.

In the AСD triangle, the OH segment is its middle line, then OH = AD / 2.

Let the length of AD= 2 * X cm, then OH = AD / 2 = X cm.

In a right-angled triangle KOH, Sin60 = OH / KН.

KN = OH / Sin60 = X / (√3 / 2) = 2 * X / √3 cm.

The area of ​​the СDK triangle is: Ssdk = СD * KН / 2 = 8 * √3.

2 * X * (2 * X / √3) / 2 = 8 * √3.

2 * X ^ 2 = 24.

X ^ 2 = 12.

X = 2 * √3 cm.

AD = СD = 2 * X = 4 * √3 cm.

KH = 2 * 2 * √3 / √3 = 4 cm.

Then Savsd = AD2 = 48 cm2.

OH = AD / 2 = 4 * √3 / 2 = 2 * √3 cm.

In a right-angled triangle, KOH, KO2 = KH2 – OH2 = 16 – 12 = 4.

KO = 2 cm.

Vpyr = Sbn * KO / 3 = 48 * 2/3 = 32 cm3.

Answer: The volume of the pyramid is 32 cm3.



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