If the perimeter of a rectangle is 44 cm and the area is 117 square cm.What is the length and width?
117 = 9*13 = 3*39
so, which pair of factors adds up to 22? (perimeter = 2*22)
Let's solve this step by step.
Step 1: We know that the perimeter of a rectangle is given by the formula: 2(length + width). In this case, the perimeter is 44 cm. So we can write the equation as:
2(length + width) = 44
Step 2: We also know that the area of a rectangle is given by the formula: length * width. In this case, the area is 117 square cm. So we can write the equation as:
length * width = 117
Step 3: Now let's solve the equation from Step 1 for length:
2(length + width) = 44
length + width = 22
length = 22 - width
Step 4: Substitute the value of length from Step 3 into the equation from Step 2:
(22 - width) * width = 117
Step 5: Expand the equation:
22*width - width^2 = 117
Step 6: Rearrange the equation to a quadratic form:
width^2 - 22*width + 117 = 0
Step 7: Factorize the quadratic equation:
(width - 13)(width - 9) = 0
Step 8: Set each factor equal to zero and solve for width:
width - 13 = 0 or width - 9 = 0
width = 13 or width = 9
So, the possible widths of the rectangle are 13 cm and 9 cm.
Step 9: Substitute the values of width into the equation from Step 3 to find the corresponding lengths:
If width = 13 cm, then length = 22 - 13 = 9 cm
If width = 9 cm, then length = 22 - 9 = 13 cm
So, the possible lengths of the rectangle are 9 cm and 13 cm.
Therefore, the possible dimensions of the rectangle are 9 cm x 13 cm and 13 cm x 9 cm.
To find the length and width of the rectangle, let's assign variables. Let's say the length of the rectangle is 'l' and the width is 'w'.
We know that the perimeter of a rectangle is given by the formula:
Perimeter = 2 * (length + width)
Since the perimeter is given as 44 cm, we can write the equation as:
44 = 2 * (l + w) (equation 1)
We also know that the area of a rectangle is given by the formula:
Area = length * width
Since the area is given as 117 square cm, we can write:
117 = l * w (equation 2)
Now, we have two equations with two variables. We can solve these equations simultaneously to find the values of 'l' and 'w'.
To eliminate one variable, we can multiply equation 1 by 'l' and equation 2 by 2:
44l = 2lw (equation 3)
234 = 2lw (equation 4)
Since equation 3 and equation 4 are equal, we can equate them:
44l = 234
Dividing both sides by 44, we get:
l = 234/44 = 5.32 (rounded to two decimal places)
Now, substitute the value of 'l' into equation 2:
117 = 5.32 * w
Dividing both sides by 5.32, we get:
w = 117/5.32 = 22 (rounded to the nearest whole number)
Therefore, the length of the rectangle is approximately 5.32 cm and the width is 22 cm.
You know that perimeter is L + L + w + w and that is equal to 44
so your first equation is 44 = 2L + 2w
and your second equation is Area and your formula is L x W
So 117 = L x W is your second equation.
44 = 2L + 2w is your other equation : )
Now depending on the grade you are in you have multiple options for solving.
Option 1: re-arrange one of your equations for a variable (say L), then sub this into the other equation and solve for w.
This is called "substitution" if that rings a bell : )
Does that option help?