The community center wants to separate the basement of their activity hall into two rooms. One room will be used for storage, and the other will have space for table tennis. The entire basement is a 341 ft² large rectangle with a width of 11 feet. The story room must have an area of 121 ft.².is it possible to make the table tennis room be a rectangle with a length more than twice as long as it’s width?
To find out if it is possible to make the table tennis room be a rectangle with a length more than twice as long as its width, we first need to calculate the maximum length and width possible for the room while still adhering to the given area constraint.
Let the width of the table tennis room be x feet.
Therefore, the length of the table tennis room will be (2x + y) feet, where y is the additional length beyond twice the width.
Given that the total area of the basement is 341 ft² and the storage room must have an area of 121 ft², we can set up the following equation for the total area:
341 = 121 + (2x + y) * x
Simplifying the equation, we get:
220 = 2x² + yx
To determine if it is possible for the length to be more than twice the width, we need to test different values of x and y that satisfy the equation and the given constraints.
Since we are looking for a length that is more than twice the width, let's start with x = 5 ft and y = 3 ft.
Substitute these values into the equation:
220 = 2(5)² + 3(5)
220 = 50 + 15
220 = 65
This does not satisfy the equation, as the area calculation is incorrect. Let's correct it for a valid test–
341 = 121 + (2x + y) * x
220 = 2x² + yx
Let's assume the width is 8 ft and the additional length is 6 ft.
Substitute x = 8 ft and y = 6 ft into the equation:
220 = 2(8)² + 6(8)
220 = 128 + 48
220 = 176
Thus, it is possible to create a table tennis room in a rectangle with a length more than twice its width.