Solution:
Given:
A frustrum with a rectangular top and base.
To find the volume of the figure, we make the sketch as shown below.
The volume of the frustrum is given by;
[tex]Volume\text{ of frustum}=volume\text{ of big pyramid}-volume\text{ of small pyramid}[/tex]
The volume of a pyramid is given by;
[tex]\begin{gathered} V=\frac{1}{3}Ah \\ where; \\ A\text{ is the base area} \\ h\text{ is the height} \end{gathered}[/tex]
Hence, we get the heights of both pyramids by similarity;
[tex]\begin{gathered} height\text{ of big pyramid}=25.7+x \\ height\text{ of small pyramid}=x \\ base\text{ of small pyramid}=15 \\ base\text{ of big pyramid}=37 \\ \\ Hence,\text{ by similarity,} \\ \frac{h_1}{b_1}=\frac{h_2}{b_2} \\ \frac{25.7+x}{37}=\frac{x}{15} \\ \\ Cross\text{ multiplying,} \\ 15\left(25.7+x\right)=37\times x \\ 385.5+15x=37x \\ Collecting\text{ the like terms,} \\ 385.5=37x-15x \\ 385.5=22x \\ Dividing\text{ both sides by 22;} \\ \frac{385.5}{22}=x \\ x=17.52cm \\ \\ Thus, \\ he\imaginaryI ght\text{ of b}\imaginaryI\text{g pyram}\imaginaryI\text{d}=25.7+x=25.7+17.52 \\ he\mathrm{i}ght\text{ of b}\mathrm{i}\text{g pyram}\mathrm{i}\text{d}=43.22cm \\ \\ he\mathrm{i}ght\text{ of small pyram}\mathrm{i}\text{d}=x \\ he\mathrm{i}ght\text{ of small pyram}\mathrm{i}\text{d}=17.52cm \end{gathered}[/tex]
Hence, the volume of the pyramids is calculated;
[tex]\begin{gathered} V=\frac{1}{3}Ah \\ Volume\text{ of the big pyramid}=\frac{1}{3}\times l\times b\times h \\ Volume\text{ of the big pyramid}=\frac{1}{3}\times37\times20\times43.22=31,982.8cm^3 \\ \\ Volume\text{ of the small pyram}\imaginaryI\text{d}=\frac{1}{3}\times l\times b\times h \\ Volume\text{ of the small pyram}\imaginaryI\text{d}=\frac{1}{3}\times15\times20\times17.52=5256cm^3 \\ \\ \\ Volume\text{ of the figure}=31982.8-5256 \\ Volume\text{ of the figure}=26726.80cm^3 \end{gathered}[/tex]
Therefore, the volume of the figure to the nearest hundredth is 26,726.80 cubic centimeters.
