Answer:
Manuel's argument is correct. Paul used the incirrect base area to find the volume of square pyramid X
Step-by-step explanation:
step 1
Find the volume of the cone W
we know that
The volume of the cone if given by the formula
[tex]V=\frac{1}{3}\pi r^{2} h[/tex]
we have
[tex]r=6\ cm\\h=5\ cm[/tex]
substitute the given values
[tex]V=\frac{1}{3}\pi (6)^{2} (5)[/tex]
[tex]V=60\pi\ cm^3[/tex]
assume
[tex]\pi=3.14[/tex]
substitute
[tex]V=60(3.14)=188.4\ cm^3[/tex]
step 2
Find the volume of the square pyramid
we know that
The volume of the pyramid is given by the formula
[tex]V=\frac{1}{3}Bh[/tex]
where
B is the area of the base
h is the height of pyramid
In this problem we have that
[tex]B=\pi r^2[/tex] ----> is the same that the area of the base of cone
so
[tex]B=3.14(6^2)=113.04\ cm^2[/tex]
[tex]h=5\ cm[/tex] ----> is the same that the height of the cone
so
substitute
[tex]V=\frac{1}{3}(113.04)(5)=188.4\ cm^3[/tex]
therefore
Manuel's argument is correct. Paul used the incirrect base area to find the volume of square pyramid X