Answer:
[tex]h=8\frac{1}{2}\text{ inch}[/tex]
Step-by-step explanation:
Let h represent height of prism.
We have been given that Tana fills the prism with base lengths [tex]3\frac{1}{4}[/tex] inch and [tex]4[/tex] inch with a [tex]110\frac{1}{2}\text{ inch}^3[/tex].
We know that volume of prism is area of base times height, so we can set an equation as:
[tex]\text{Volume of prism}=\text{Base length}\times\text{Base width}\times \text{Height of prism}[/tex]
[tex]110\frac{1}{2}\text{ inch}^3=3\frac{1}{4}\text{ inch}\times 4\text{ inch}\times h[/tex]
[tex]\frac{221}{2}\text{ inch}^3=\frac{13}{4}\text{ inch}\times 4\text{ inch}\times h[/tex]
[tex]\frac{221}{2}\text{ inch}^3=13\text{ inch}^2\times h[/tex]
[tex]\frac{221}{2*13\text{ inch}^2}\text{ inch}^3=\frac{13\text{ inch}^2\times h}{13\text{ inch}^2}[/tex]
[tex]\frac{221}{26}\text{ inch}=h[/tex]
[tex]8\frac{13}{26}\text{ inch}=h[/tex]
[tex]8\frac{1}{2}\text{ inch}=h[/tex]
Therefore, the height of the prism is [tex]8\frac{1}{2}\text{ inch}[/tex].
