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QUESTION: A conference room table is to be constructed, rectangular in shape, with semicircles at the ends.

Perimeter of table- 40 feet

Area of rectangular portion- twice sum of areas of two ends

Calculate length and with of rectangular part.

Thanks.

table
ANSWER: The rectangular portion is W feet wide and L feet long.
area of rectangular portion = WL

The semicircles stretch across the width of the table. Combined, they form a circle of radius = W/2.
area of circle = π(W/2)² = πW²/4
circumference = πW

area of rectangle = 2(area of circle)
WL = πW²/2
L = πW/2

perimeter of table = 2L + πW
2L + πW = 40
2(πW/2) + πW = 40
2πW = 40
W = 20/π ≅ 6.37
L = πW/2 = π(20/π)/2 = 10

The rectangular part is 10 feet long and approximately 6.37 feet wide.

Check:
circumference of circle = 6.37π ≅20 ft
perimeter of table = 2·10 + 20 = 40 ft

area of rectangle ≅ 10·6.37 = 63.7 ft²
area of circle = π(6.37/2)² = 31.87 ft²
2(area of circle) ≅ 63.7 = area of rectangle.

---------- FOLLOW-UP ----------

QUESTION: "The semicircles stretch across the width of the table. Combined, they form a circle of radius = W/2.
area of circle = π(W/2)² = πW²/4
circumference = πW"

I don't understand this part of your solution.

Put the semicircles together to make a whole circle.
Since the width of the circle is the same as the width of the table,

area of circle = πr² = π(W/2)² = πW²/4
circumference of circle = 2πr = 2π(W/2) = πW
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