IS THIS EQUATION IS DIMENSIONALLY CORRECT?
P= L + W where P is the perimeter, L is the length, and W is the width.
Times the area by the perimeter and that will give you the answer.
Length of rectangle is 3 cm more than twice its width, area=35 cm2
To find the area and perimeter of a rectangle you need to know formulas for each:
Area of a rectangle=L*W
Perimeter of a rectangle=2(L+W)
So to find either the area or the perimeter of a rectangle you need to know to things.
The value of:
The rectangles length
The rectangles width
To illustrate lets suppose that the length of the rectangle is 2 and the width is 3. Hence:
Width equal 7 feet less than its length and area is 18 square feet?
find the dimensions of the rectangle
All rectangles of a given area will have the same perimeter,.Therefore knowing both will not help you to determine the dimensions.
Draw a graph of a rectangle with length 1', Width 10', Perimeter 22', area 10'sq
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I dunt get this please help
It is not true that all rectangles of a given area will have the same perimeter. To prove this to yourself, imagine a rectangle measuring 2 x 2, with a perimeter of 8, and area of 4. Then compare to a rectangle measuring 1 x 4, with a perimeter of 10, but with the same area of 4. To solve the problem of your question, you will need to do some algebra, as follows:
Perimeter = P = 2 (Length) + 2(Width) = 2L + 2W
Area = A = (Length)(Width) = LW
Rearrange the equation for perimeter to find W:
P = 2L + 2W = 2( L + W)
(P/2) = L + W
W = (P/2) - L
If you substitute this value into the Area equation,
A = LW = L( (P/2) - L) = (PL/2)-(L^2)
A = -L^2 + (PL/2)
Or, rearranging into standard form,
Now you can use the quadratic formula to solve for L.
Remember that for an equation in standard form,
aX^2 + bX + c = 0
x = (-b +/- sqrt(b^2 - 4ac)) / 2a
So our equation (where x is L) has
a = 1
b = (P/2)
c = A
L = (-(P/2) +/- sqrt((P/2)^2 - 4A)) / 2
Now, given perimeter and area, you can solve for the dimensions. As an example, P = 15, A = 14
L = (-(15/2) +/- sqrt((15/2)^2 - 4*14)) / 2
L = (-7.5 +/- .5) / 2 = -8 / 2 OR -7 / 2
L = 4 OR 3.5
(You can change the length to a positive number because when measuring length, positive or negative just depends on the direction)
You get 2 answers due to the way the quadratic formula works. However, it makes sense, because you would get two different answers depending on whether the Length or Width of your rectangle was the long dimension. So, choosing L=4, the dimensions of the recangle would be 4 x 3.5 .
As a check,
A = LW = (4)(3.5) = 14
p = 2L + 2W = 2(4) + 2(3.5) = 8 + 7 = 15
You can write a quadratic equation and solve it.
area = L*W
perimeter = 2(L+W)
area = L*(perimeter/2 - L) (substitute for W)
L2 - (perimeter/2)*L + area = 0 (rearrange to standard form)
L = (perimeter/2 ±√((perimeter/2)2 - 4(area)))/2 (use the quadratic formula)
L = perimeter/4 ± √(perimeter2/16 - area) (simplify)
Please note that there are two possible values for L. If you were to solve for W, you would get the same equation. This means that W is one of the values and L is the other. If you want L > W, then you will choose
L = perimeter/4 + √((perimeter/4)2 - area)
W = perimeter/4 - √((perimeter/4)2 - area)
Suppose we know the perimeter of a 2x3 plot is 10, and its area is 6. Our formulas give
L = 10/4 + √((10/4)2-6) = 2.5 + √(6.25-6) = 2.5+.5 = 3
W = 2.5 - .5 = 2
You will note that L = W = perimeter/4 = √area if the figure is a square.
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