The area of a rectangle is 24, and its diagonal is radical 52. Find its dimensions and perimeter.
Longer side = ?
Shorter side = ?
Perimeter = ?
Let the dimensions be a and b.
ab = 24
a^2 + b^2 = 52
a^2 + (24/a)^2 = 52
a^4 -52a^2 + 576 = 0
(a^2 -36)(a^2 -16)
a^2 = 36 or 16
a = 6 or 4
b = 4 or 6
Thank You!
Let X and Y represent the sides of
the rectangle.
X^2 + Y^2 = 52 = diagonal squared.
XY = 24 =the area of rectangle.
y = 24/X
X^2 + (24/X)^2 = 52
X^2 +576/X^2 = 52
Multiply both sides by x^2 :
X^4 + 576 = 52X^2
X^4 - 52X^2 + 576 = 0. used computer to
help find roots: X = -4, X = 4, X = -6,
X = 6. Select the positive roots:
The sides are 4 and 6. P = (2*4) +(2*6)
= 20
To find the dimensions and perimeter of the rectangle, we need to use the given information.
Let's assume the longer side of the rectangle is denoted as "l" and the shorter side as "w."
We are given two pieces of information:
1. The area of the rectangle is 24, which means: l * w = 24.
2. The diagonal of the rectangle is √52.
To proceed, let's use the Pythagorean theorem. According to this theorem, in a rectangle, the length of the diagonal (d) can be expressed as:
d = √(l^2 + w^2)
Since we are given that the diagonal is √52, we can write the equation as:
√52 = √(l^2 + w^2)
Squaring both sides of the equation, we get:
52 = l^2 + w^2 ... (Equation 1)
Now, we have two equations:
l * w = 24 ... (Equation 2)
l^2 + w^2 = 52 ... (Equation 1)
From Equation 2, we can solve for one variable in terms of the other:
w = 24 / l
Substituting this value of w into Equation 1, we have:
l^2 + (24 / l)^2 = 52
Expanding and simplifying, we get:
l^4 - 52l^2 + 24^2 = 0
Now we have a quadratic equation in l^2. Let's solve for l^2 using the quadratic formula:
l^2 = [ -(-52) ± √((-52)^2 - 4(1)(24^2)) ] / 2(1)
l^2 = (52 ± √(2704 - 2304)) / 2
l^2 = (52 ± √400) / 2
l^2 = (52 ± 20) / 2
Solving for the two possible values of l^2, we get:
l^2 = 36 or l^2 = 16
Taking the square root of both sides, we get:
l = 6 or l = -4
Since lengths cannot be negative, we discard the negative value of l.
So, the longer side (l) is 6.
Now, substituting this value back into Equation 2, we can solve for the shorter side (w):
6w = 24
w = 24 / 6
w = 4
Hence, the dimensions of the rectangle are:
Longer side = 6
Shorter side = 4
To find the perimeter, we use the formula: Perimeter = 2(l + w)
Substituting the values we found, we get:
Perimeter = 2(6 + 4)
Perimeter = 2 × 10
Perimeter = 20
Therefore, the perimeter of the rectangle is 20 units.