find thevsurface area of a square pyramid wuth side length 3 km and slant height 5 km
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The Total Surface Area = 4 triangles + 1 square
The TSA of the Pyramid = 4(1/2 bh) + LxL
[tex]\begin{gathered} d^2=3^2+3^2 \\ d^2=9+9 \\ d=\sqrt[]{18}\text{ =}\sqrt[]{9\times2}=\sqrt[]{9}\text{ }\times\sqrt[]{2}\text{ =3 }\sqrt[]{2} \end{gathered}[/tex][tex]\begin{gathered} 5^2=h^2+(\frac{3}{2})^2 \\ 25-\frac{9}{4}=h^2 \\ 25-2.25=h^2 \\ h=\sqrt[]{22.5}\text{ =1.5km} \end{gathered}[/tex]TSA of the pyramid =
[tex]4(\frac{1}{2}\times5\times3)+(3^2)=(2\times15)+9=30+9=39km^2[/tex]