Question

All the edges of a square pyramid are length $12$ centimetre.
(a) What is the area of one lateral face of it?
(b) What is the surface area of this pyramid?
(c) Prove that the square of any term of this sequence is also a term of it.

Answer 1

(a) Since all the edges of a square pyramid are of length $12cm$, lateral faces of the pyramid are equilateral triangles having each edge of $12cm$.
Thus, area of one lateral face $=\frac{\sqrt{3}}{4}\times 12\times 12=36\sqrt{3}sq.cm$
(b) An edge $=12cm=a$
Slant height $=l=\frac{\sqrt{3}}{2}\times 1=6\sqrt{3}cm$
Surface area of pyramid
$=a^2+4\times \frac{\sqrt{3}}{4}{a}^2$
$={12}^2+\frac{\sqrt{3}}{4}\times {12}^2$
$=144\times 144\sqrt{3}$
$=144(1+\sqrt{3})sq.cm$
(c) Length of edge of pyramid, $a=12cm$
New length of edge, $a^{\prime }=24cm$
New surface area of the pyramid
$=(a^{\prime }{)}^2+4\times \frac{\sqrt{3}}{4}(a^{\prime }{)}^2$
$={24}^2+4\times \frac{\sqrt{3}}{4}\times {24}^2$
$=576+576\sqrt{3}$
$=576(1+\sqrt{3})sq.cm$
Previous surface are, $S=144(1+\sqrt{3})sq.cm$
New surface area, $S^{\prime }=576(1+\sqrt{3})sq.cm$
$\Rightarrow S^{\prime }=4S$
Thus, if the length of the sides of the pyramid is doubled, the surface area will be $4$ times the previous one.