Question

Can someone help me on this?? I'm stuck!

Find the total area for the regular pyramid.

T. A. =

Answer 1

Answer:

$TA=(144+36\sqrt{3})\ units^2$

Step-by-step explanation:

we know that

The total area or surface area of the regular pyramid is equal to the area of the triangular base plus the area of its three lateral triangular faces

so

step 1

Find the area of the triangular base B

Is an equilateral triangle

Applying the law of sines

$B=\frac{1}{2}(12^2)sin(60^o)$

$B=\frac{1}{2}(144)\frac{\sqrt{3}}{2}$

$B=36\sqrt{3}\ units^2$

step 2

Find the area of the lateral triangular faces

$A=3[\frac{1}{2}(12)h]$

Find the height

Applying the Pythagorean Theorem

$10^2=6^2+h^2$

$h^2=100-36\\h^2=64\\h=8\ units$

Find the area of the lateral triangular faces

$A=3[\frac{1}{2}(12)8]=144\ units^2$

therefore

The total area is

$TA=(144+36\sqrt{3})\ units^2$