The diagram shows a solid prism ABCDEF whose cross section is an isosceles triangle. It is given that $AF = 9\ \text{cm}$, $BC = 15\ \text{cm}$ and $\angle AFB = 36^\circ$.

1. Find the area of the triangular cross-section $\triangle AFB$.

Since $AF=FB=9\text{ cm}$ and $\angle AFB=36^\circ$,
$\text{Area of }\triangle AFB=\frac12(9)(9)\sin36^\circ$
$=23.8059\ldots\text{ cm}^2$

2. Multiply the cross-sectional area by the prism length to get the volume.

$\text{Volume}=\text{area of cross-section}\times \text{length}$
$=23.8059\ldots\times 15$
$=357.088\ldots\text{ cm}^3$

1. Find the volume of one sphere of radius $0.8$ cm.

Volume of one sphere
$=\frac{4}{3}\pi r^3$
$=\frac{4}{3}\pi(0.8)^3$
$=2.14466\ldots\text{ cm}^3$

2. Divide the prism volume by the volume of one sphere.

Using the volume of the prism from part (a),
$\text{number of spheres}=\frac{356}{2.14466\ldots}$
$=165.99\ldots$

3. Compare the calculated number with Johnny's claim.

$165.99\ldots \approx 166$
Therefore Johnny's claim is correct.