A 5cm piece of wire is cut into two parts such that a square formed by bending on part will have four times the area of a square formed by bending the other part, what is the length of the longer part.
let the piece formed into the first square be x cm
then the piece formed into the second square is 5-x cm
area of first square = (x/4)^2 = x^2/16
area of 2nd square = (5-x)^2/16
4x^2/16 = (5-x)^2/16
4x^2 = 25 - 10x + x^2
3x^2 + 10x - 25 = 0
(3x-5)(x+5) = 0
x = 5/3 or x = -5, the last part makes no sense
so one part is 5/3 cm
the other part is 5 - 5/3 or 10/3 cm
check:
area of smaller is (5/12)^2 = 25/144
area of larger is (10/12)^2 = 100/144 , which is 4 times as large.
Let's denote the length of the longer part as x cm.
According to the given information, the other part will have a length of (5 - x) cm.
When the longer part is bent into a square, each side of the square will have a length of x/4 cm, and the area of the square will be (x/4)^2 = x^2/16 square cm.
When the other part is bent into a square, each side of the square will have a length of (5-x)/4 cm, and the area of the square will be ((5-x)/4)^2 = (25 - 10x + x^2)/16 square cm.
It is given that the square formed by the longer part will have four times the area of the square formed by the other part. So we can equate the two areas:
x^2/16 = 4 * (25 - 10x + x^2)/16
Multiplying both sides of the equation by 16, we get:
x^2 = 4 * (25 - 10x + x^2)
Expanding the right side of the equation, we have:
x^2 = 100 - 40x + 4x^2
Rearranging the terms, we have:
3x^2 - 40x + 100 = 0
This is a quadratic equation. We can solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 3, b = -40, and c = 100. Substituting these values into the quadratic formula, we get:
x = (-(-40) ± √((-40)^2 - 4 * 3 * 100)) / (2 * 3)
x = (40 ± √(1600 - 1200)) / 6
x = (40 ± √400) / 6
Now, we can simplify further:
x = (40 ± 20) / 6
Using the positive root:
x = (40 + 20) / 6
x = 60 / 6
x = 10 cm
Therefore, the length of the longer part is 10 cm.
To solve this problem, let's break it down step by step:
Let's assume the length of the shorter part of the wire is x cm. This means the length of the longer part is (5 - x) cm.
When we bend the shorter part of the wire to form a square, the length of each side of the square will be x/4 cm. Therefore, the area of this square will be (x/4)^2 cm^2.
When we bend the longer part of the wire to form a square, the length of each side of the square will be (5 - x)/4 cm. Therefore, the area of this square will be ((5 - x)/4)^2 cm^2.
According to the problem, the square formed by bending the longer part of the wire should have four times the area of the square formed by bending the shorter part. So we can write the equation:
((5 - x)/4)^2 = 4 * (x/4)^2
Simplifying this equation:
((5 - x)/4)^2 = x^2/4
Cross multiplying:
(5 - x)^2 = x^2
Expanding:
25 - 10x + x^2 = x^2
Subtracting x^2 from both sides:
25 - 10x = 0
Rearranging the equation:
10x = 25
x = 25/10
x = 2.5 cm
Since x is the length of the shorter part of the wire, the length of the longer part will be:
5 - 2.5 = 2.5 cm
Therefore, the length of the longer part is 2.5 cm.