The document describes constructing a conference table with a rectangular center portion and semicircular ends. The perimeter is 40 feet and the area of the rectangle is twice the sum of the semicircular areas. To find the rectangle dimensions: 1) The semicircles form a full circle of radius r; 2) The rectangle area equals twice the circular area, giving an equation to solve for r; 3) The perimeter equation with r solved for gives r = 10/pi; 4) Solving for the rectangle length L using r and the perimeter gives L = 10 feet.

1. A large table for a conference Room is to be constructed in the shape of a rectangle with two
semi circles at the ends. The table has a perimeter of 40 feet, and the area of the rectangle portion
is to be twice the sum of the areas of the two ends. Find the length and width of the rectangular
portion.
Solution
The two semicircles of ends make up one circle.
The area of the circle is pi*r^2
The area of the rectangle in the center is twice the area of the circle:
Area = 2 * pi*r^2
The Area of the rectangle is also equal to 2 * the radius of the circle time L, with L being the
side
not adjacent to one of the ends of the semi circle.
L * 2*r = 2 * pi * r^2
Solve for L:
L = pi * r
Now lets write an equation for the perimeter which is 40 feet long:
2*pi *r + 2*L = 40 Area of a circle + the 2 sides that are not adjacent to the semicirles.
Substitute for L from above:
2*pi*r + 2*pi*r = 40
Solve for r yields 10/pi.
Remember L = pi * r, plug in the value for r to get L:
L = pi * 10/pi the pi's cancel and L = 10
Dimentions of rectange: 10 X (20/pi)