Question

Find thevsurface area of a square pyramid wuth side length 3 km and slant height 5 km

Answer 1

The Total Surface Area = 4 triangles + 1 square

The TSA of the Pyramid = 4(1/2 bh) + LxL

`\begin{gathered} d^2=3^2+3^2 \\ d^2=9+9 \\ d=\sqrt[]{18}\text{ =}\sqrt[]{9*2}=\sqrt[]{9}\text{ }*\sqrt[]{2}\text{ =3 }\sqrt[]{2} \end{gathered}`
`\begin{gathered} 5^2=h^2+((3)/(2))^2 \\ 25-(9)/(4)=h^2 \\ 25-2.25=h^2 \\ h=\sqrt[]{22.5}\text{ =1.5km} \end{gathered}`

TSA of the pyramid =

`4((1)/(2)*5*3)+(3^2)=(2*15)+9=30+9=39km^2`