What is the area of a rectangle whose length is 1 unit longer than its width and whose diagonal is 2 unit longer than the width?
looks like Pythagoras.
width --- x
length --- x+1
diagonal --- x+2
x^2 + (x+1)^2 = (x+2)^2
by recognition:
the sides are one unit apart, the only Pythagorean triple like that is 3,4,5, since 3^2 + 4^2 = 5^2.
If you did not recognize that:
x^2 + x^2 + 2x + 1 = x^2 + 4x + 4
x^2 - 2x - 3 = 0
(x - 3)(x+1) = 0
x = 3 or x = -1, but a side can't be negative, so
x = 3
find the area using base*height
To find the area of the rectangle, we need to determine both its length and width. Let's start by assigning variables. Let's say the width of the rectangle is 'w' units.
According to the given information:
- The length of the rectangle is 1 unit longer than its width. So, the length would be 'w + 1' units.
- The diagonal is 2 units longer than the width. Using the Pythagorean theorem, we know that the diagonal, the length, and the width form a right triangle. The diagonal can be calculated as √(length² + width²). In this case, the diagonal is 'w + 2' units.
Now, using the Pythagorean theorem, we can express the equation as:
(length)² + (width)² = (diagonal)²
Substituting the values we have:
(w + 1)² + w² = (w + 2)²
Expanding and simplifying:
w² + 2w + 1 + w² = w² + 4w + 4
Combining like terms and canceling out w² from both sides:
2w + 1 = 4w + 4
Simplifying further:
1 - 4 = 4w - 2w
-3 = 2w
Dividing both sides by 2:
w = -3/2
However, since width cannot be negative, we have a contradiction. Therefore, there is no valid solution for this problem.