the length of a rectangle is 23 centimeters less than five times its width. Its area is 42 square centimeters. Find the dimensions of the rectangle
the length of a rectangle is 23 centimeters less than five times its width. Its area is 42 square centimeters. Find the dimensions of the rectangle
Well, well, well, we have ourselves a geometry problem! Let's solve it with a touch of humor, shall we?
Let's say the width of the rectangle is "W" (because using "X" wouldn't be as wide-reaching). According to the problem, the length is 23 centimeters less than five times the width, which we can express as 5W - 23 (because why be plain when you can be fancy?).
Now, to find the area, we simply multiply the width (W) by the length (5W - 23). And voila! We have our equation: W * (5W - 23) = 42.
To make this equation a little happier, let's simplify it: 5W^2 - 23W - 42 = 0. Now we can solve it using quadratic equations, factoring, or even asking nicely. I'll use factoring because it makes me feel happy.
We can factor it as (W - 7)(5W + 6) = 0. Which means either W - 7 = 0 or 5W + 6 = 0.
If W - 7 = 0, then W = 7. But let's not assume too quickly, my friend!
If 5W + 6 = 0, then subtracting 6 from both sides gives us 5W = -6. Divide both sides by 5, and we find W = -6/5.
But wait a second – we're talking about a rectangle's dimensions here! And negative values just won't cut it. So let's stick with W = 7.
Now that we have the width (W = 7), we can find the length using our earlier expression: 5W - 23. Plugging in W = 7 gives us 5(7) - 23 = 35 - 23 = 12.
Ta-da! The dimensions of the rectangle are a width of 7 centimeters and a length of 12 centimeters. Just remember, it's not just about the answer, it's about the journey – and the laughter along the way!
To solve this problem, we can set up two equations based on the given information.
Let's represent the width of the rectangle as "w" and the length as "l".
From the given information, we can determine the following:
1) "The length of a rectangle is 23 centimeters less than five times its width":
This can be represented as: l = 5w - 23
2) "Its area is 42 square centimeters":
The area of a rectangle is given by: A = l * w
Substituting the values, we get: 42 = (5w - 23) * w
Now we have a quadratic equation: 5w^2 - 23w - 42 = 0
To solve this equation, we can either factor it or use the quadratic formula.
For simplicity, let's use the quadratic formula:
w = (-b ± sqrt(b^2 - 4ac))/(2a)
Here, a = 5, b = -23, c = -42.
Substituting the values, we get:
w = (-(-23) ± sqrt((-23)^2 - 4 * 5 * (-42))) / (2 * 5)
w = (23 ± sqrt(529 + 840)) / 10
w = (23 ± sqrt(1369)) / 10
w = (23 ± 37) / 10
Simplifying further:
w = (23 + 37) / 10 = 60 / 10 = 6
OR
w = (23 - 37) / 10 = -14 / 10 = -1.4
Since the width of a rectangle cannot be negative, we can disregard the negative value.
So, the width of the rectangle is 6 cm.
Now, we can use this value to find the length of the rectangle:
l = 5w - 23
l = 5 * 6 - 23
l = 30 - 23
l = 7 cm
Therefore, the dimensions of the rectangle are:
Width = 6 cm, Length = 7 cm.
w(5w-23) = 42
Now solve for w.
Or, consider that two factors of 42 are 6 and 7.