hello
given that the pyramid has the shape of a triangle, we can easily find the height of the pyramid using pythagoran's theorem
from triangle b, let's use the formula and solve for y
[tex]\begin{gathered} x^2=h^2+z^2 \\ 6^2=h^2+1.5^2 \\ 36=h^2+2.25 \\ \text{collect like terms} \\ h^2=36-2.25 \\ h^2=33.75 \\ \text{solve for h} \\ h=\sqrt[]{33.75} \\ h=5.809\approx5.81m \end{gathered}[/tex]
having known the value of the heigh of the pyramid, we can now proceed to solve for the lateral area and surface area
for the lateral area, the formula is given as
[tex]\begin{gathered} A_l=l\sqrt[]{l^2+4h} \\ l=\text{edge length} \\ h=\text{height of pyramid} \end{gathered}[/tex][tex]\begin{gathered} A_l=l\sqrt[]{l^2+4h} \\ l=3m \\ h=5.81m \\ A_l=3\sqrt[]{3^2+4\times5.81_{}} \\ A_l=17.03m^2 \end{gathered}[/tex]
the lateral area of the figure is 17.03 squared meter.
let's solve for the surface area
the formula for the surface area of a square pyramid is given as
[tex]\begin{gathered} A=l^2+2l\sqrt[]{\frac{l^2}{4}+4h^2} \\ l=3m \\ h=5.81 \\ A=3^2+2\times3\sqrt[]{\frac{3^2}{4}+4\times5.81^2} \\ A=9+6\sqrt[]{\frac{9}{4}+135.0244} \\ A=79.298\approx79.3m \end{gathered}[/tex]
