Bricks to Cover a SteepleDate: 08/23/97 at 15:43:29 From: Bailey Rector Subject: Bricks to cover a steeple Our Literature teacher asked us to figure out how many bricks it would take to cover the steeple of a Roman Church in Florence. The steeple is an octagon with a diameter of 130 feet and a height of 370 feet. Each brick is 6 inches by 12 inches. How do I start to solve this problem? Date: 08/29/97 at 13:21:05 From: Doctor Rob Subject: Re: Bricks to cover a steeple First you need to know the side length of an octagon with diameter 130 feet. Draw a one-eighth part of an octagon with radii from the centerto the vertices. You get an isosceles triangle whose vertex angle is 45 degrees and whose base angles are 67.5 degrees each, and whose equal sides have length 65 feet. Use either the law of sines or the law of cosines to compute the length of the base s. For example, using the law of cosines, s^2 = 65^2 + 65^2 - 2*65*65*cos(45 degrees) = 4225 + 4225 - 8450*Sqrt[2]/2 = 65^2*(2 - Sqrt[2]) s = 65*Sqrt[2 - Sqrt[2]] = 49.7488462 Then the area of each face of the octagonal prism is 370*s square feet and the lateral area of the octagonal prism will be 8*370*s square feet. Each brick accounts for 1/2 square foot area, so the number of bricks is 8*370*s/(1/2) = 2*8*370*s, at least. Since this is not an integer, you will have to round upwards. The number of bricks is at least this number, if there is no waste at the corners. If you insist that no brick surface bend around a corner, you will have to round s up and use 50*370*8*2 for the number of bricks, which is somewhat larger. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/30/97 at 12:06:13 From: Rick Rector Subject: Re: Bricks to cover a steeple Dear Dr Rob, Thank you for your response. My dad helped me with the problem and we came up with the same answer on the sides of the octagon. We did not have the same answer to the rest. We found the sides of the inside triangle, which is the height of the center of the steeple of 370 and goes out 65 from the center of the octogon at the base. This gave us a right triangle with sides 370 and 65. Using the Pythagorean theorem we squared both numbers and got 136900 + 4225 = 141125, which has a square root of 375.66607. Dad said this should be the length of the cornor of one side of the prism. Then to find the area of one side of the prism we found the height. We figured the side of the prism was a triangle that had a base of 49.7488462 and two equal sides of 375.66607. This was cut in half to get a right triangle with a base of 24.874423 and hypotenuse of 375.66607. The square of these are 618.73691 and 141125, so 141125 - 618.73691 = 140506.26309, which has a square root of 374.84165 - which gave us a triangle with a base of 49.7488462 and a height of 374.84165. With area of 1/2bh .5*49.7488462*374.84165 = 9323.9697 square feet on each side times 8 sides totalled to 74591.757 square feet. The 6" by 12" brick is 1/2 of a square foot, which meant we need two bricks per square foot. So we multiplied 74591.757 times 2 to get 149183.51 bricks or about 149200 bricks. I turned this in and my teacher said it was not the answer or even close. He did not give us the answer yet. Inside triangle in steeple ^ Side of steeple /| / \ / | / | \ 375.6 / | / | \ 375.6 / | 370 / | \ / | /374.8 \ / | / | \ /------| /------|------\ 65 49.75 What are we doing wrong? Thank you for your help Dr Rob. Sincerely, Bailey Rector Date: 09/04/97 at 15:08:22 From: Doctor Rob Subject: Re: Bricks to cover a steeple First, I think I know what I did wrong. I assumed from your statement that the brick part of the steeple was the lateral surface of an octagonal prism, that is, the upper and lower faces were congruent regular octagons with diameter 130 feet, and the sides were all rectangles 370 feet high. Apparently the steeple is, in fact, an octagonal pyramid, with a point, and whose single base is the regular octagon with diameter 130 feet. The area you seek is apparently the area of eight congruent isosceles triangles whose bases have length equal to the side of the regular octagon. You are correct that the length of the equal sides of the isosceles triangles is Sqrt[141125] = 375.666075, and the altitude of each triangle is 374.841651 feet. The area is 9323.969857 square feet for each triangle, or 74,591.758514 square feet altogether. Each brick is 1/2 square foot, so 149183.517028 bricks would be needed, or 149,184, as you indicated. It is possible that there is a miscommunication between you (and me) and the teacher. The "height" of 370 feet may mean the height of the triangles making up the lateral surface of the pyramid, as opposed to the perpendicular height from the vertex to the base. In that case, the area would be 8*(1/2)*370*49.7488462 = 73628.292376, and the number of bricks twice that, or 147256.584752, rounded up to 147,257, or about 147,300. That would account for the different answers found. I hope this helps, and thanks for straightening me out about the prism vs. pyramid mixup. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/