Volume Calculation for an Icosidodecahedron

If you extend the faces of all the pentagons of the icosidodecahedron, it will form a dodecahedron where each vertex is centered above one of the equilateral triangles.

A pentagon in the icosidodecahedron can be formed by joining the midpoints of a pentagon in the dodecahedron. Since the icosidodecahedron pentagon has sides of $1$ , the dodecahedron pentagon has sides of $s = 2 \cdot \frac{1}{2} \sec 36° = \frac{1}{2}(\sqrt{5} - 1)$ . That makes the volume of the dodecahedron:

$V_{\text{dodec}} = \frac{1}{4}(15 + 7\sqrt{5})s^3 = \frac{1}{4}(15 + 7\sqrt{5})(\frac{1}{2}(\sqrt{5} - 1))^3 = 10 + 2\sqrt{5}$

Each of the "extra" $20$ pyramids formed by the difference of the dodecahedron and icosidodecahedron have equilateral triangle bases with sides of $1$ and lateral edges of $l = \frac{1}{2}s = \frac{1}{4}(\sqrt{5} - 1)$ . The radius of the equilateral triangle base is $r = \frac{1}{\sqrt{3}}$ , so by the Pythagorean Theorem the height of the pyramid is $h = \sqrt{l^2 - r^2} = \sqrt{(\frac{1}{4}(\sqrt{5} - 1))^2 - (\frac{1}{\sqrt{3}})^2} = \frac{1}{6}(3\sqrt{3} - \sqrt{15})$ . That makes the volume of each pyramid:

$V_{\text{pyr}} = \frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot 1^2 \cdot \frac{1}{6}(3\sqrt{3} - \sqrt{15}) = \frac{1}{24}(3 - \sqrt{5})$

Since the volume of the icosidodecahedron is the volume of a dodecahedron minus the volume of $20$ pyramids with equilateral triangle bases:

$V_{\text{ico}} = V_{\text{dodec}} - 20 \cdot V_{\text{pyr}} = 10 + 2\sqrt{5} - 20 \cdot \frac{1}{24}(3 - \sqrt{5}) = \frac{1}{6}(45 + 17\sqrt{5}) \approx \boxed{13.8355}$

For a regular n-gon with side 1, the distance from centre to a vertex is given by $r_n=\frac{1}{2 \sin π/n}$ and its area is given by $A_n=\frac{n}{4\tan π/n}$

If you look carefully you can see that between opposite pentagons there is a regular 10-gon dividing the solid in half. This means that the distance from the centre to any vertex equals $r_{10}$ and the distance from the centre to each of the faces can be calculated using pythagoras: $h_n=\sqrt{r_{10}^2-r_n^2}$

The volume of the pyramid that a face has to the centre is given by $V_n=\frac{1}{3}A_nh_n$

Now the total volume is the sum of 32 such pyramids: $12 V_5+20V_3\approx 13.83552594$

I used the shortcut of looking up the formula of icosahedron, and from the question deduced that the original one before the trimming is size 2. Then I just deducted the volume of 12 pentamids from the results.
Formula for volume of icosahedron.
= 5(3 + √5)s³ / 12.
V_icos(2) = (30 + 10√5) / 3.

Volume of a unit of icosahedral vertices' uniform pentamid of size 1.
= (5 + √5) / 24.

Volume of icosidodecahedron.
= V_icos(2) - 12 x [(5 + √5) / 24].
= (45 + 17√5) /6