Finding the length and the width?

Question

A large table for a conference room is to be constructed in the shape of a rectangle with two semicircles at the ends. The table is to have a perimeter of 40 ft, and the area of the rectangular portion is to be twice the sum of the areas of the two ends. Find the length and the width of the rectangular portion.

Answer 1

The required length is 10 ft, and the required width is 20/π ft.



If you analyze the appearance of the table, it will look like this: (top only)

http://welcome.com.au/shop/images/d%20end%20table.JPG



Hence, the width of the rectangle is not included in calculating the perimeter. The most effective method to solve this problem is to use a system of equations to represent the given information.



First equation:

2L + πW = 40



With the πW, remember that the perimeter of a circle is its circumference. (C = πD). However, we have a semicircle, and the formula for its circumference is (C = 1/2(πD)). Then again, since we have two semi-circles, and their circumferences are included in the perimeter, it's as if we have a circle, nevertheless. Their diameter is the width of the rectangle, so we have (1/2)πW + (1/2)πW = πW.



Second equation:

LW = πW² / 2



The problem states that the area of the rectangle is twice the area of the two semi-circles, or the circle. Remember that our diameter is the width of the rectangle, and since we need the radius to find the area, we need half of the width. Hence, we have 2π(W/2)² = πW² / 2.



Eliminate one of the variables to solve for the other.



2L + πW = 40

LW - πW² / 2 = 0



2L + πW = 40

2LW - πW² = 0



W(2L + πW = 40)

2LW - πW² = 0



2LW + πW² = 40W

2LW - πW² = 0



4LW = 40W

L = 10 ft



Finding the width:

2L + πW = 40

2(10) + πW = 40

πW = 40 - 20

πW = 20

W = 20/π ft



Dimensions: l = 10 ft; w = 20/π ft



Hope this helps!

Answer 2

you know:

the circumference of the table is 40 ft. it consists of two semicircles and two sides of the rectangle.



2 * pi * r + 2 * a = 40

2 * (pi * r + a) = 40

pi * r + a = 20



the area of the rectangle ( one side is a, the other side is 2 * the radius of the semicircles ) is:

A = a * 2r



the area of the semicircles is:

A = pi * r^2



twice the area of both semicircles equals the area of the rectangle.



a * 2 * r = 2 * ( pi * r^2 )

2 * a * r = 2 * pi * r^2

a * r = pi * r^2

a = pi * r



insert this expression for "a" into the term you got for the circumference:



pi * r + a = 20

pi * r + ( pi * r ) = 20

2 * pi * r = 20

pi * r = 10

r = 10 / pi

r = 3.18309886...



you know that a = pi * r, so:



a = pi * r

insert the value r = 10/pi

a = pi * 10 / pi

a = 10



this is the length of the rectangular portion.



the width is 2 * r.



2 * r = 2 * 3.18309886 = 6.36619772



answer: the length of the rectangular portion is 10 ft, the width is 6.366 ft.

Answer 3

xy = (pi * (x/2)^2)*2...area of rectangle = (xy), and (pi * (x/2)^2)*2 is two times the area of the ends



pi*x+2y=40.....perimeter



solve simultaneous equations gets the following answers, which checks OK



x = 20 / pi

y = 10



y is the length of the rectangular portion

x is it width..which is also the diameter of the ends



good luck