Question

10 POINTS
Find the surface area of the regular pyramid. Round to the nearest tenth if necessary.

Answer 1

Answer:

$SA=234.8\ yd^2$

Step-by-step explanation:

we know that

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

step 1

Find the area of the triangular base

we know that

The triangular base is an equilateral triangle

so

The area applying the law of sines is equal to

$A=\frac{1}{2}(14^2)sin(60^o)$

$A=\frac{1}{2}(196)\frac{\sqrt{3}}{2}$

$A=49\sqrt{3}=84.87\ yd^2$            

step 2

Find the area of its three triangular lateral faces

$A=3[\frac{1}{2}bh]$

we have

$b=14\ yd$

Find the height of triangles

Applying the Pythagorean Theorem

$10^2=(14/2)^2+h^2$

solve for h

$100=49+h^2$

$h^2=100-49$

$h=\sqrt{51}\ yd$

substitute

$A=3[\frac{1}{2}(14)\sqrt{51}]$      

$A=149.97\ yd^2$

step 3

Find the surface area

Adds the areas

$SA=84.87+149.97=234.84\ yd^2$

Round to the nearest tenth

$SA=234.8\ yd^2$

Answer 2

Answer:

234.8 is correct.

Step-by-step explanation: