U'll have to draw this image to really understand the following explanation. If 10 is the height then 16 and 12 are the Length and Breadth.
Imagine a box with dimensions 16,12,10. Now, picture this box from the top i.e. u see a flat 16*12 rectangle. The lamp is in the very middle of this rectangle. Using Pythagoras, we can figure that the diagonals of this rectangle =√(16²+12²)=20 .So the center, is at a distance of 10 from each vertex of this rectangle. Now, the longest distance, that we are required to find, from this center point will be at any of the vertices at the bottom of this box.
Now, switch to 3D view. Project what we saw on the ceiling onto the floor. As the ceiling and floor have the same dimensions. U have a vertical line of length 10 which joins the lamp to the center of the floor at the bottom. Lets call this point X. Point X is equidistant from all the bottom vertices. And the distance from X to any bottom vertex is 10(Simply project the information from the ceiling onto the floor).
So, we have a right triangle with perpendicular sides 10 and 10 and we need to find the Hypotenuse. Using Pythagoras, √(10²+10²)= 10√2.........which is the farthest distance betn the lamp and any point in the room.
geometry
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sud21 wrote:A lamp is in the center of the ceiling of a room with dimension 16, 12, 10, where 10 is the height. What is the longest distance between the lamp and any point in the room?
In the figure above, d is the longest distance between the lamp and any point in the room.
Since the lamp is in the center of the room -- halfway across the length of 16 and the width of 12 -- d is the MAIN DIAGONAL of a rectangular solid whose dimensions are l=8, w=6, and h=10.
The formula for the main diagonal of a rectangular solid is d = √(l² + w² + h²). (Some people call this formula the SUPER-PYTHAGOREAN theorem.)
Thus, in the figure above:
d = √(8² + 6² + 10²) = √200 = 10√2.
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