PERIMETER OF A RECTANGLE IS 32CM AND AREA IS 55CM FIND THE LENGHT.
l+w = 16 --> l=16-w
lw=55
w(16-w)=55
16w - w^2=55
w^2-16w+55=0
(w-5)(w-11)=0w=5 or w=11
we have a symmetric solution, that is,
if w=5 , then l=11
if w=11, then l = 5
the rectangle is 5 by 11
usually the longer side is called the length
so the length is 11 cm
To find the length of a rectangle when the perimeter and area are given, we can use the following formulas:
1. Perimeter of a rectangle = 2(l + w), where l is the length and w is the width.
2. Area of a rectangle = l × w.
Given that the perimeter is 32 cm and the area is 55 cm², we can set up two equations:
Equation 1: 32 = 2(l + w)
Equation 2: 55 = l × w
Let's solve these equations step by step:
Step 1: Solve Equation 1 for w:
32 = 2(l + w)
Divide both sides by 2:
16 = l + w
Subtract l from both sides:
16 - l = w
Step 2: Substitute the value of w from Equation 1 into Equation 2:
55 = l × w
Substitute w = 16 - l:
55 = l × (16 - l)
Step 3: Expand Equation 3 and arrange it in standard form:
55 = 16l - l²
Step 4: Rearrange the equation to form a quadratic equation:
l² - 16l + 55 = 0
Step 5: Factorize the quadratic equation, if possible. If not, use the quadratic formula. However, in this case, the equation doesn't factorize nicely. So let's solve it using the quadratic formula:
The quadratic formula states that for an equation ax² + bx + c = 0, the solutions for x are given by the following formula:
x = (-b ± √(b² - 4ac)) / (2a)
Applying this formula to our equation:
l = [16 ± √(16² - 4(1)(55))] / 2
l = [16 ± √(256 - 220)] / 2
l = [16 ± √36] / 2
l = (16 ± 6) / 2
Step 6: Solve for l:
l = (16 + 6) / 2 = 22 / 2 = 11
l = (16 - 6) / 2 = 10 / 2 = 5
So, the length of the rectangle can be either 5 cm or 11 cm, depending on the orientation of the rectangle.