Right pyramid with square base volume surface and angles

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A right pyramid with square base $ABCD$ and apex $V$ has base edge $8$ and lateral-face apothem $5$.

(a) Compute the volume.

(b) Compute the total surface area.

(c) Find the angle between the lateral faces and the base.

(d) Find the angle $A\hat BV$.

(e) Find the radius of the inscribed sphere (resting on the base and tangent to the lateral faces).

Solution

The base apothem (half the side) is $4$, so the height of the pyramid is $h=\sqrt{5^2-4^2}=3$.

(a) $V=\tfrac13\cdot64\cdot3=64$.

(b) $S=64+4\cdot\left(\tfrac12\cdot8\cdot5\right)=64+80=144$.

(c) The dihedral angle between a lateral face and the base has tangent $\tfrac{h}{4}=\tfrac34$, so $\varphi\approx36.87°$.

(d) The lateral edge is $\overline{VB}=\sqrt{h^2+(4\sqrt2)^2}=\sqrt{41}\approx6.403$. In the isosceles triangle $ABV$ (with $\overline{AB}=8$ and $\overline{VA}=\overline{VB}=\sqrt{41}$), by the law of cosines: $\cos(A\hat BV)=\dfrac{\overline{AB}^2+\overline{VB}^2-\overline{VA}^2}{2\,\overline{AB}\,\overline{VB}}=\dfrac{64}{16\sqrt{41}}=\dfrac{4}{\sqrt{41}}\approx0.625\Rightarrow A\hat BV\approx51.34°$.

(e) The inscribed sphere, tangent to all faces, has radius $r=\dfrac{3V}{S}=\dfrac{192}{144}=\dfrac43\approx1.333$. $\boxed{V=64;\ S=144;\ \varphi\approx36.87°;\ A\hat BV\approx51.34°;\ r=\tfrac43\approx1.33}$