To determine which option is correct, we first need to find the volume of both chocolate.
The volume of X:
The shape is a square-based pyramid. The volume is given by
[tex]\begin{gathered} V_x=\frac{1}{3}\times base\text{ area }\times height \\ V_x=\frac{1}{3}\times l\times b\times h \end{gathered}[/tex]
From the diagram,
l = 5 cm
b = 6 cm
h = 10 cm
Substituting,
[tex]\begin{gathered} V_x=\frac{1}{3}\times5\times6\times10 \\ V_x=100\operatorname{cm}^3 \end{gathered}[/tex]
The volume of Y:
The shape is a triangular-based pyramid. The volume is given by
[tex]\begin{gathered} V_y=\frac{1}{3}\times base\text{ area }\times height \\ V_y=\frac{1}{3}\times\frac{1}{2}\times b\times l\times h \end{gathered}[/tex]
From the diagram,
l = 8 cm
b = 7.5 cm
h = 10 cm
[tex]\begin{gathered} V_y=\frac{1}{3}\times\frac{1}{2}\times7.5\times8\times10 \\ V_y=100\operatorname{cm}^3 \end{gathered}[/tex]
From here, the volumes of both chocolates are the same.
Therefore, the chocolate he picks does not matter as both volumes are equivalent.
The SECOND OPTION is correct.
