pythagoras theorem
First one
[tex]hypotenuse^2=base^2+perpendicular[tex]10^2=5^2+h^2\\h^2=5^2=10^2\\h^2=25-100\\h^2=75\\h=\sqrt{75} \\h=8.7[/tex]^2[/tex]
c) 8.7m
second one
the surface area of the pyramid
[tex]SA=a^{2}+2a\sqrt{\frac{a^2}{4}+h^2}[/tex]
[tex]12^2+2(12)\sqrt{\frac{12^2}{4}+11^2} \\144+24\sqrt{36+121}\\168\sqrt{157}\\444.71[/tex]
c)528sq ft
