SOLUTION: You decide to build a box that is 4 feet long ,5 feet wide, and 8 feet high. The distance from the top left front corner to the top righ tback corner is ? feet. The distance from

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: You decide to build a box that is 4 feet long ,5 feet wide, and 8 feet high. The distance from the top left front corner to the top righ tback corner is ? feet. The distance from       Log On


   

Question 13785: You decide to build a box that is 4 feet long ,5 feet wide, and 8 feet high. The distance from the top left front corner to the top righ tback corner is ? feet.
The distance from the bottom left front corner to the top right back corner is ? feet.
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
1) Let the distance from the top-left front corner to the top-right back corner = d1.
The top is a 4 ft by 5 ft rectangle and d1 is a diagonal of the top.
Use the Pythagorean theorem to find the length of d1. d1 is the hypotenuse of the right triangle whose sides the length of the box (4 ft.) and the width of the box (5 ft.)
d1+=+sqrt%284%5E2+%2B+5%5E2%29
d1+=+sqrt%2816+%2B+25%29
d1+=+sqrt%2841%29feet.
d1 = 6.4 ft. (approximately)
2) Let the distance from the bottom-left front corner to the top-right back corner = d2
d2 is a diagonal of the rectagular prism (the box) and its length can be found using the Pythagorean theorem. d2 is the hypotenuse of the right triangle whose sides are the diagonal of the top of the box, d1 =(sqrt%2841%29 ft.) and the height of the box (8 ft.)
d2+=+sqrt%288%5E2+%2B+d1%5E2%29
d2+=+sqrt%2864+%2B+sqrt%2841%29%5E2%29
d2+=+sqrt%2864+%2B+41%29
d2+=+sqrt%28105%29 feet.
d2 = 10.25 ft. (approximately)