the diagonal of a rectangular room is 52ft long. one wall measures 28ft longer than the ajacent wall. find the dimensions of the room
Let x be the length of the shorter side, then
x²+(x+28)² = 52²
Solve for x.
Hint: x is a positive integer.
To find the dimensions of the room, let's assign variables to the lengths of the adjacent walls.
Let's say the length of one wall is x feet. Since the other wall is 28 feet longer than the adjacent wall, we can represent it as (x + 28) feet.
Now, we can use the Pythagorean theorem to relate the lengths of the walls to the diagonal. According to the theorem, in a right-angled triangle, the square of the length of the hypotenuse (diagonal) is equal to the sum of the squares of the other two sides (walls).
In this case, we have the following equation:
x^2 + (x + 28)^2 = 52^2
Let's solve this equation to find the value of x.
Expanding and simplifying the equation:
x^2 + x^2 + 56x + 784 = 2704
Combining like terms:
2x^2 + 56x + 784 = 2704
Moving all terms to one side to form a quadratic equation:
2x^2 + 56x + 784 - 2704 = 0
Simplifying further:
2x^2 + 56x - 1920 = 0
Now we can solve this quadratic equation to get the values of x.
We can either factor this equation or use the quadratic formula. Let's use the latter.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = 56, and c = -1920.
Plugging these values into the quadratic formula, we get:
x = (-56 ± √(56^2 - 4*2*(-1920))) / (2*2)
Simplifying:
x = (-56 ± √(3136 + 15360)) / 4
x = (-56 ± √18496) / 4
x = (-56 ± 136) / 4
Now, we have two possibilities:
1. x = (-56 + 136) / 4 = 80 / 4 = 20
2. x = (-56 - 136) / 4 = -192 / 4 = -48
Since we cannot have negative lengths for the walls, we discard the second solution (-48) and consider the first solution (x = 20).
Now, we can find the length of the other wall (x + 28):
20 + 28 = 48
Therefore, the dimensions of the room are:
Length = 20 ft
Width = 48 ft