Question

The base of a right pyramid is an equilateral triangle each side of which is $6\sqrt{3}$ cm long and its height is 4 cm Find the total surface area of the pyramid in $c{m}^2$

• A. $64\sqrt{3}$
• B. $68\sqrt{3}$
• C. $72\sqrt{3}$
• D. $76\sqrt{3}$

Answer 1

The correct option is C $72\sqrt{3}$

$\tan 30°=\frac{l}{3\sqrt{3}};l=3$

Slant height $h=\sqrt{3^2+4^2}=5$

Area of one triangle face $=\frac{1}{2}\times 6\sqrt{3}\times 5$

$=15\sqrt{3}$

Lateral S.A $=3\times 15\sqrt{3}$

$=45\sqrt{3}c{m}^2$

Area of base $=\frac{\sqrt{3}}{4}{a}^2\Rightarrow \frac{\sqrt{3}}{4}\times {\left(6\sqrt{3}\right)}^2=27\sqrt{3}$

Total area $=45\sqrt{3}+27\sqrt{3}=72\sqrt{3}c{m}^2$