Surface area of a pyramid = [tex]lw+l \sqrt{ \frac{w^2}{2}+h^2 } +w \sqrt{ \frac{ l^{2} }{2} + h^{2} } [/tex]
[tex]84=6
\times 6 + 6( \sqrt{ \frac{6^2}{2} +h^2}) +6( \sqrt{ \frac{6^2}{2}
+h^2}) \\ 84=36+12( \sqrt{ \frac{36}{2} +h^2}) \\ 48=12( \sqrt{ 18
+h^2}) \\ 4= \sqrt{18 +h^2} \\ 16=18 +h^2 \\ h^2=16-18=-2 \\ h=
\sqrt{-2} [/tex]
Which is not real. Hence, the question is not correct.
