Answer:
![SA=234.8\ yd^2](https://tex.z-dn.net/?f=SA%3D234.8%5C%20yd%5E2)
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)](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%2814%5E2%29sin%2860%5Eo%29)
![A=\frac{1}{2}(196)\frac{\sqrt{3}}{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%28196%29%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D)
step 2
Find the area of its three triangular lateral faces
![A=3[\frac{1}{2}bh]](https://tex.z-dn.net/?f=A%3D3%5B%5Cfrac%7B1%7D%7B2%7Dbh%5D)
we have
![b=14\ yd](https://tex.z-dn.net/?f=b%3D14%5C%20yd)
Find the height of triangles
Applying the Pythagorean Theorem
![10^2=(14/2)^2+h^2](https://tex.z-dn.net/?f=10%5E2%3D%2814%2F2%29%5E2%2Bh%5E2)
solve for h
![100=49+h^2](https://tex.z-dn.net/?f=100%3D49%2Bh%5E2)
![h^2=100-49](https://tex.z-dn.net/?f=h%5E2%3D100-49)
![h=\sqrt{51}\ yd](https://tex.z-dn.net/?f=h%3D%5Csqrt%7B51%7D%5C%20yd)
substitute
![A=149.97\ yd^2](https://tex.z-dn.net/?f=A%3D149.97%5C%20yd%5E2)
step 3
Find the surface area
Adds the areas
![SA=84.87+149.97=234.84\ yd^2](https://tex.z-dn.net/?f=SA%3D84.87%2B149.97%3D234.84%5C%20yd%5E2)
Round to the nearest tenth
![SA=234.8\ yd^2](https://tex.z-dn.net/?f=SA%3D234.8%5C%20yd%5E2)
The answer would be A. 3.6 × 10^9
Answer:
A cylinder has parallel discs bases that are congruent in size.
Step-by-step explanation:
Step-by-step explanation:
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