find volumes of cubes: for A, L=H=w=10_5/8; for B, L=H=w=21

[downandshady]

The problem says

The welded tanks A and B are made from a 1/8 in steel plate. Outside measurments are given.

A 's L is 10 5/8'' H is 10 5/8'' and the w is 10 5/8''

B 's L is 21'' H is 21'' and w is 21''

find the volume of tank A. Round the answer to the nearest tenth cubic inch.

find the volume of tank B.

 

[soroban]

Re: I cant seem to figure it out

Hello, downandshady!

Are you waiting for some magic formula?

\(\displaystyle \text{The welded tanks }A\text{ and }B\text{ are made from }\tfrac{1}{8}\text{-inch steel plate.}\)

\(\displaystyle \text{They are both cubes. }\:\text{Outside measurments are given.}\)

\(\displaystyle \text{Cube A: }\:10\tfrac{5}{8}\text{ inches on an edge.}\)'

\(\displaystyle \text{Cube B: }\:21\text{ inches on an edge.}\)


\(\displaystyle \text{Find the volume of the tanks. }\:\text{Round to the nearest tenth cubic inch.}\)

\(\displaystyle \text{Do to the thickness of the metal, the inner cube is a }\tfrac{1}{4}\text{-inch smaller on an edge.}\)

\(\displaystyle \text{The interior of Cube A is a cube which is }10\tfrac{3}{8}\text{ inches on an edge.}\)

\(\displaystyle \text{Its volume is: }\10\tfrac{3}{8})^3 \:=\\tfrac{83}{8})^3 \:=\:1116.771484 \;\approx\;1116.8\text{ in}^3\)


Now you try Cube B . . .