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The surface area of a hexoganal pyramid with lateral edges of 13cm and base edges of 10cm?

You have 6 triangles that are 13x13x10cm

And you have the regular hexagon pyramid base with each side length 10cm.

The area of each of the triangles is given by

\(A_{face} = b h = b \sqrt{s^2-\dfrac{b^2}{4}}cm^2\)

there are 6 of these so the total area of the faces is 

\(A_{sides}=6b\sqrt{s^2-\dfrac {b^2}{4}}cm^2\)

Now the area of the hexagonal base is given by

\(A_{base}=\dfrac{3\sqrt{3}}{2}b^2cm^2\)

plugging the values for b and s in we get

\(A = A_{sides}+A_{base} = 6(10)\sqrt{(13)^2-\dfrac{(10)^2}{4}}+\dfrac{3\sqrt{3}}{2}(10)^2=\\ \\ 720+150 \sqrt{3}cm^2\)