The solid object shown is the frustum of a square based pyramid..
To calculate the total surface area, we shall follow the procedure below;
[tex]\begin{gathered} \text{Total Surface Area}=TSA \\ \text{TSA}=\lbrack\frac{1}{2}(P_1+P_2)L\rbrack+B_1+B_2 \\ \text{Where the variables are;} \\ P_1=Perimeter\text{ of base 1} \\ P_2=\text{Perimeter of base 2} \\ L=Slant\text{ height} \\ B_1=\text{Area of base 1} \\ B_2=\text{Area of base 2} \end{gathered}[/tex]We shall now solve as follows;
[tex]\begin{gathered} P_1=2(l+w) \\ P_1=2(14+14) \\ P_1=2(28) \\ P_1=56in \end{gathered}[/tex][tex]\begin{gathered} P_2=2(l+w) \\ P_2=2(24+24) \\ P_2=2(48) \\ P_2=96in \end{gathered}[/tex][tex]\begin{gathered} B_1=l\times w \\ B_1=14\times14 \\ B_1=196in^2 \end{gathered}[/tex][tex]\begin{gathered} B_2=l\times w \\ B_2=24\times24 \\ B_2=576in^2 \end{gathered}[/tex]The total surface area would now be;
[tex]\begin{gathered} \text{TSA}=\lbrack\frac{1}{2}(P_1+P_2)L\rbrack+B_1+B_2 \\ \text{TSA}=\lbrack\frac{1}{2}(56+96)19\rbrack+196+576 \\ \text{TSA}=\lbrack\frac{1}{2}(152)19\rbrack+772 \\ \text{TSA}=\lbrack76\times19\rbrack+772 \\ \text{TSA}=1444+772 \\ \text{TSA}=2216in^2 \end{gathered}[/tex]The volume of a frustum of a square based pyramid is given as;
[tex]\begin{gathered} Vol=\frac{1}{3}h(B_1+B_2+\sqrt[]{B_1B_2}) \\ \text{Where;} \\ h=vertical\text{ height (altitude)} \end{gathered}[/tex]The volume would now be;
[tex]\begin{gathered} \text{Vol}=\frac{1}{3}\times15(196+576+\sqrt[]{196\times576}) \\ \text{Vol}=5(772+\sqrt[]{112896}) \\ \text{Vol}=5(772+336) \\ \text{Vol}=5(1108) \\ \text{Vol}=5540in^3 \end{gathered}[/tex]ANSWER:
Therefore, we have
Total surface area = 2,216 inches squared
Volume = 5,540 inches cubed
