SOLUTION: The lateral edge of a pyramidal church spire is 92 ft. Each side of its octagonal base is 43 ft.What will be the cost of painting the spire at Php 530.00 a square foot?
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Question 1194335: The lateral edge of a pyramidal church spire is 92 ft. Each side of its octagonal base is 43 ft.What will be the cost of painting the spire at Php 530.00 a square foot?
You can put this solution on YOUR website! **1. Find the Perimeter of the Base**
* Perimeter of the octagonal base = Number of sides * Length of each side
* Perimeter = 8 sides * 43 ft/side = 344 ft
**2. Calculate the Lateral Surface Area**
* Lateral Surface Area = (1/2) * Perimeter of base * Slant height
* Lateral Surface Area = (1/2) * 344 ft * 92 ft
* Lateral Surface Area = 15824 sq ft
**3. Calculate the Cost of Painting**
* Cost = Lateral Surface Area * Cost per square foot
* Cost = 15824 sq ft * Php 530.00/sq ft
* Cost = Php 8,392,320.00
**Therefore, the cost of painting the spire will be Php 8,392,320.00**
You can put this solution on YOUR website! .
The lateral edge of a pyramidal church spire is 92 ft.
Each side of its octagonal base is 43 ft.
What will be the cost of painting the spire at Php 530.00 a square foot?
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The solution in the post by @proyaop is FATALLY INCORRECT.
It is incorrect, since @proyaop mistakenly treats the lateral edge of the pyramid
as its slant height, although (and despite the fact) that they are different conceptions.
I came to bring a correct solution.
The perimeter of the pyramidal church spire is p = 43 ft * 8 = 344 ft.
The slant height is h = = 89.4525 ft.
The lateral area of the spire is = = 15385.83 ft^2.
The cost is 530*15385.83 = Php 8154489.90. ANSWER
Solved.
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To check, I calculated the area of one single lateral face as the area of the (92,92,43)-ft
isosceles triangle using the Heron's formula.
I used online calculator https://www.mathopenref.com/heronsformula.html
It gave me the area of 1923.2288 ft^2 for one single triangle.
For 8 triangles, it gives the total lateral area 1923.2288*8 = 15385.83 ft^2,
which is the same as in my calculations above.