Question

a square pyramid has a base height edge length of 3m and a slant height of 6m. find the lateral area and surface area of the pyramid

Answer 1

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

`\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}`

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

`\begin{gathered} A_l=l\sqrt[]{l^2+4h} \\ l=\text{edge length} \\ h=\text{height of pyramid} \end{gathered}`
`\begin{gathered} A_l=l\sqrt[]{l^2+4h} \\ l=3m \\ h=5.81m \\ A_l=3\sqrt[]{3^2+4*5.81_{}} \\ A_l=17.03m^2 \end{gathered}`

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

`\begin{gathered} A=l^2+2l\sqrt[]{(l^2)/(4)+4h^2} \\ l=3m \\ h=5.81 \\ A=3^2+2*3\sqrt[]{(3^2)/(4)+4*5.81^2} \\ A=9+6\sqrt[]{(9)/(4)+135.0244} \\ A=79.298\approx79.3m \end{gathered}`