SOLUTION: Some houses, like the one shown at the right, have roofs that are in the shape of right pyramids with rectangular bases. (It's just a picture of a regular house). The base of the f

Algebra ->  Surface-area -> SOLUTION: Some houses, like the one shown at the right, have roofs that are in the shape of right pyramids with rectangular bases. (It's just a picture of a regular house). The base of the f      Log On


   

Question 956157: Some houses, like the one shown at the right, have roofs that are in the shape of right pyramids with rectangular bases. (It's just a picture of a regular house). The base of the front part of the house shown in the photo is rectangular, 20 ft by 30 ft. The walls of the house are 9ft height and the peak of the roof is 14 ft high. Find the surface area of the roof. I can't use the formula (1/2)(l)(p) + B, because the base is irregular. I tried the Pythagorean Theorem, to find the area of the lateral faces and then added the base to it, but it didn't get me to the answer I was looking for. The answer is 651.638 ft squared. A thank you for anyone who helps!!!
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
The height of the pyramid is 14ft-9ft=5ft
Half the length of the 20 ft side is one leg to find the slant height of the 30 ft side, the other leg is the height.
sqrt%2810%5Eft%5E2%2B5%5E2ft%5E2%29=sqrt%28125ft%5E2%29=slant height 30ft side
The two triangles that form the roof on the two 30 ft sides have area
2%281%2F2%29%2830ft%29sqrt%28125ft%5E2%29=30%28sqrt%28125ft%5E2%29%29=335.410ft%5E2
Half the length of the 30 ft side is one leg to find the slant height of the 20 ft side, the other leg is the height.
sqrt%2815%5E2ft%5E2%2B5%5E2ft%5E2%29=sqrt%28250ft%5E2%29%29=slant height 20ft side.
The two triangles that form the roof on the two 20 ft sides have area
2%281%2F2%29%2820ft%29sqrt%28250ft%5E2%29=20%28sqrt%28250ft%5E2%29%29=316.228ft%5E2
Total area of the roof=335.410 sq ft+316.228 sq ft=651.638 sq ft.