The area of the given square pyramid is:
total area = 1,100 inches squared.
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How to get the area of the pyramid?</h3>
On the second image, we can see that the pyramid is conformed of a square base and 3 triangles.
To get the surface area of the pyramid, we can just get the area of each of these simpler parts.
The base is a square of 22 in by 22 in, then the area of the base is:
B = (22in)*(22 in) = 484 in^2
For each triangle, the area will be:
A = (base side)*(height)/2
A = (22in)*(14in)/2 = 154 in^2
And we have 4 of these triangles, then the total area of the pyramid will be:
total area = B + 4*A = 484in^2 + 4*(154 in^2) = 1,100 in^2
If you want to learn more about square pyramids:
brainly.com/question/22744289
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Answer:A triangle has a perimeter of 165 cm. The first side is 65 cm less than twice the second side. The third side is 10 cm less than the second side. Write and solve an equation to find the length of each side of the triangle.
The sides can be found by taking the square root of the area. , where = side. . So the length of a side is 6.3 cm.
Step-by-step explanation:
So, we know the center is at -1, -3, hmmm what's the radius anyway?
well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -1 &,& -3~) % (c,d) &&(~ -7 &,& -5~) \end{array} \\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2} \\\\\\ r=\sqrt{36+4}\implies r=\sqrt{40}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-1%20%26%2C%26%20-3~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%20-7%20%26%2C%26%20-5~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B%5B-7-%28-1%29%5D%5E2%2B%5B-5-%28-3%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%28-7%2B1%29%5E2%2B%28-5%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B36%2B4%7D%5Cimplies%20r%3D%5Csqrt%7B40%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{40}}{ r} \\\\\\\ [x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B-1%7D%7B%20h%7D%2C%5Cstackrel%7B-3%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%5Csqrt%7B40%7D%7D%7B%20r%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-1%29%5D%5E2%2B%5By-%28-3%29%5D%5E2%3D%28%5Csqrt%7B40%7D%29%5E2%5Cimplies%20%28x%2B1%29%5E2%2B%28y%2B3%29%5E2%3D40)
All we need to do is start with the volume formula. Then, input the volume and solve the equation for the radius. And double it to get the diameter.
V = (4/3)pi(r^3)
12114.7 = (4/3)pi(r^3) Divide both sides by (4/3) and pi
2893.64 = r^3 Take the cube root of both sides
14.25 = r
So the diameter is about 28.5 cm (14.25 x 2 = 28.5)
