A rectangle,3cm longer than it is wide, has a diagonal 15cm long. What are the dimensions of the rectangle?
w^2 + (w+3)^2 = 15
remember your 3-4-5 right triangle?
multiply that by 3.
22.5
To solve this problem, let's denote the width of the rectangle as x cm. Since the length is 3 cm longer than the width, the length can be represented as (x + 3) cm.
We know that the diagonal of a rectangle forms a right triangle with the length and width as its sides. Using the Pythagorean theorem, we can find the relationship between the sides of the triangle.
According to the Pythagorean theorem:
(diagonal)^2 = (length)^2 + (width)^2
Let's substitute the given values:
(15 cm)^2 = (x + 3 cm)^2 + (x cm)^2
Simplifying the equation:
225 cm^2 = x^2 + 6x + 9 + x^2
Combining like terms:
225 cm^2 = 2x^2 + 6x + 9
Rearranging the equation to form a quadratic equation:
2x^2 + 6x + 9 - 225 = 0
2x^2 + 6x - 216 = 0
Now, we can solve this quadratic equation either by factoring, using the quadratic formula, or completing the square. Let's use factoring in this case.
Factoring:
2(x + 18)(x - 6) = 0
Setting each factor to zero and solving for x:
x + 18 = 0 or x - 6 = 0
x = -18 or x = 6
Since the width cannot be negative, we discard x = -18. Therefore, the width of the rectangle is x = 6 cm.
Now, let's find the length of the rectangle:
Length = Width + 3 = 6 cm + 3 cm = 9 cm
The dimensions of the rectangle are 6 cm by 9 cm.
To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this case, the rectangle's diagonal forms the hypotenuse of a right triangle, and the two sides of the right triangle are the width and length of the rectangle.
Let's assign variables:
Let w be the width of the rectangle.
Since the rectangle is 3cm longer than it is wide, the length of the rectangle can be expressed as w + 3.
Using the Pythagorean theorem, we have:
(w)^2 + (w + 3)^2 = (15)^2
Expanding the equation:
w^2 + (w^2 + 6w + 9) = 225
Combining like terms:
2w^2 + 6w + 9 = 225
Rearranging the equation to standard quadratic form:
2w^2 + 6w - 216 = 0
Factoring out common factors:
(w + 18)(2w - 12) = 0
Setting each factor equal to zero and solving for w:
w + 18 = 0 or 2w - 12 = 0
w = -18 or w = 6
We discard the negative value as width cannot be negative, so we have:
w = 6
Since the width of the rectangle is 6cm, the length of the rectangle is:
w + 3 = 6 + 3 = 9 cm
So, the dimensions of the rectangle are 6 cm (width) and 9 cm (length).