note: figure not drawn to scale\nthe figure shown is a right rectangular pyramid, where ( l = 16 ) units, (…

Answer:

Mathematics: Geometry
The surface area of a right rectangular pyramid is calculated by adding the area of the base and the areas of the four triangular faces. The base area is ( l\times w = 16\times8 = 128 ). The area of the two triangular faces with base ( l = 16 ) and height ( 10 ) (using Pythagoras ( \sqrt{18^{2}-(8 / 2)^{2}}=\sqrt{324 - 16}=\sqrt{308}\approx17.55 ) is wrong, assume slant height for ( l = 16 ) is ( 10 ) (from figure info if any, else using formula ( S = lw+\frac{1}{2}(2l + 2w)s ), where ( s ) is slant height. For ( w = 8 ), slant height ( s_{1}=\sqrt{18^{2}+(16/2)^{2}}=\sqrt{324 + 64}=\sqrt{388}\approx19.7 ), for ( l = 16 ), slant height ( s_{2}=\sqrt{18^{2}+(8/2)^{2}}=\sqrt{324+16}=\sqrt{340}\approx18.44 ). But if we use formula ( S= lw+\frac{1}{2}(2l\times s_{2}+2w\times s_{1}) ). Substituting ( l = 16 ), ( w = 8 ), ( s_{1}=\sqrt{18^{2}+8^{2}}=\sqrt{324 + 64}=\sqrt{388}\approx19.7 ), ( s_{2}=\sqrt{18^{2}+4^{2}}=\sqrt{324+16}=\sqrt{340}\approx18.44 ). ( S=16\times8+\frac{1}{2}(2\times16\times18.44 + 2\times8\times19.7)=128+(295.04+315.2)=738.24\approx738 ) square units. But if using wrong slant height (assuming from wrong approach in problem - like if using wrong formula ( S = lw+\frac{1}{2}(2l + 2w)h ) (wrong, but if problem has typo and assume slant height as ( h )) ( S=16\times8+\frac{1}{2}(2\times16 + 2\times8)\times18=128+(48\times9)=128 + 432 = 560 ). But correct formula ( S = lw+\frac{1}{2}(2l\times\sqrt{h^{2}+(\frac{w}{2})^{2}}+2w\times\sqrt{h^{2}+(\frac{l}{2})^{2}})=16\times8+\frac{1}{2}(2\times16\times\sqrt{18^{2}+4^{2}}+2\times8\times\sqrt{18^{2}+8^{2}})=128+(16\times\sqrt{340}+8\times\sqrt{388})\approx128+(16\times18.44+8\times19.7)=128+(295.04 + 157.6)=580.64\approx581 )