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Charra [1.4K]
3 years ago
12

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

Mathematics
2 answers:
NNADVOKAT [17]3 years ago
5 0

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

garri49 [273]3 years ago
4 0

Answer:

234.8 is correct.

Step-by-step explanation:

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