Question 1181876: The perimeter of a rectangle is 120cm. The area of the rectangle is 864cm^2. Determine the length and width of the rectangle. Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52756) (Show Source):
You can put this solution on YOUR website! .
The perimeter of a rectangle is 120cm. The area of the rectangle is 864cm^2.
Determine the length and width of the rectangle.
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The standard algebra solution is as follows.
Let w be the width length, in centimeters.
Then the length is (120/2-w) = 60-w centimeters long.
The area is w*(60-w); therefore, the "area" equation is
w*(60-w) = 864.
Write it in the standard form quadratic equation
w^2 - 60w + 864 = 0.
Use the quadratic formula
= = = = 30 +- 6.
One answer is 24 cm for the width and 36 cm for the length.
The other answer is 36 cm for the width and 24 cm for the length.
ANSWER. The dimensions of the rectangle are 36 cm and 24 cm.
The standard solution is completed.
Now I will show you a non-standard one.
The sum of the width and the length is 120/2 = 60 centimeters.
Notice that the mean of the width and the length values is the number of = 30.
This number is exactly half-way between the length and the width values.
So, the length = 30+x, while the width = 30-x centimeters, where x is some unknown value.
Since the area of the rectangle is 864 cm^2 (given), you have this equation
(30+x)*(30-x) = 864,
or
900 - x^2 = 864,
x^2 = 900 - 864
x^2 = 36
x = = 6.
Thus the dimensions of the rectangle are 30+6 = 36 cm and 30-6 = 24 cm.
You get the same answer.
Solved (in two ways, and the second way is practically MENTAL solution).
I write this response to give hearty endorsement to the method show by tutor @ikleyn for solving this problem. It is a trick that can make the solution of this and many similar problems easier -- easy enough that someone with good mental math skills can solve it in a very short time.
Let's review the standard algebraic setup for solving the problem.
The perimeter is 120, so the length plus width is 60; and the area is 864. So we divide the "length plus width" into two parts, x and (60-x).
Multiplying length times width to get the area leads us to the quadratic equation x^2-60x+864=0.
To try to solve that by factoring, we need to find two numbers whose sum is 60 and whose product is 864.
But that's what the original problem required us to do -- so the formal algebra hasn't gotten us any closer to the solution. We can of course use the quadratic formula to get the solution....
But the other tutor's setup for solving the problem makes the solution far easier.
Instead of breaking the 60 into x and (60-x), break it into (30+x) and (30-x).
That makes the equation (30+x)(30-x)=864, which is easily solved:
