A rectangle has area 6 cm (squared) and diagonal of length 2√5 cm. What is its perimeter?
thanks
let the width be x and the length be y
we know xy = 6 and
x^2 + y^2 = (2√5)^2 = 20
Perimeter = 2(x+y)
from algebra we know
(x+y)^2 = x^ + 2xy + y^2
but we know the values on the right side,
so
(x+y)^2 = 20 + 6 = 26
then x+y = √26
and the perimeter is 2√26
let the width be x and the length be y
we know xy = 6 and
x^2 + y^2 = (2�ã5)^2 = 20
Perimeter = 2(x+y)
from algebra we know
(x+y)^2 = x^ + 2xy + y^2
but we know the values on the right side,
so
(x+y)^2 = 20 +12=32
then x+y = �ã32
and the perimeter is 2�ã32
Why is math so hard?
To find the perimeter of a rectangle, we need to know both the length and width. However, in this case, we are only given the area and the diagonal length of the rectangle.
Let's start by using the formula for the area of a rectangle:
Area of a rectangle = length × width
From the given information, we have:
6 cm² = length × width
Next, let's use the Pythagorean theorem to find the relationship between the length, width, and diagonal of the rectangle. In a right-angled triangle, the square of the length of the diagonal is equal to the sum of the squares of the length and width.
Let's represent the length as "L" and the width as "W." We now have:
(L² + W²) = (2√5)² = 20 cm²
Combining this equation with the equation for the area, we have a system of two equations with two variables:
6 cm² = LW
L² + W² = 20 cm²
To solve this system, we can use substitution or elimination. Let's solve it using substitution.
Rearrange the first equation for L:
L = 6 cm² / W
Substitute this expression for L in the second equation:
(6 cm²/W)² + W² = 20 cm²
Simplify:
36 cm^4 / W^2 + W² = 20 cm²
Combine terms:
36 cm^4 + W^4 = 20W^2
Rearrange the equation:
W^4 - 20W^2 + 36 = 0
Now, we have a quadratic equation in terms of W². We can substitute z = W² to simplify the equation:
z² - 20z + 36 = 0
To solve this quadratic equation, we can use the quadratic formula:
z = (-b ± √(b² - 4ac)) / 2a
For this equation, a = 1, b = -20, and c = 36. Plugging in these values into the quadratic formula, we get:
z = (-(-20) ± √((-20)² - 4(1)(36))) / (2(1))
Simplifying further, we have:
z = (20 ± √(400 - 144)) / 2
z = (20 ± √256) / 2
z = (20 ± 16) / 2
Now, let's consider both scenarios:
When z = (20 + 16) / 2 = 18:
W² = 18 cm²
Taking the square root of both sides:
W = √18 cm = 3√2 cm
When z = (20 - 16) / 2 = 2:
W² = 2 cm²
Taking the square root of both sides:
W = √2 cm
Since we are dealing with a rectangle, we can assume that the width should be less than the diagonal. Therefore, the width is √2 cm.
Now, let's substitute the width (√2 cm) into the equation for the area to find the length:
6 cm² = L × √2 cm
Rearrange for L:
L = 6 cm² / √2 cm
Rationalize the denominator:
L = 6 cm² / √2 cm * √2 cm / √2 cm
L = 6√2 cm
Therefore, the length is 6√2 cm.
Finally, we can find the perimeter of the rectangle, which is equal to 2 times the sum of the length and width:
Perimeter = 2(L + W)
Perimeter = 2(6√2 cm + √2 cm)
Perimeter = 2(7√2 cm)
Perimeter = 14√2 cm
Hence, the perimeter of the rectangle is 14√2 cm.