SOLUTION: How do I find the perimeter of a rectangle that's in a circle when all I know is the diameter (6 inches) and the area of the square (15). I've been at this for a week please help
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Question 1190163: How do I find the perimeter of a rectangle that's in a circle when all I know is the diameter (6 inches) and the area of the square (15). I've been at this for a week please help Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52788) (Show Source):
You can put this solution on YOUR website! .
How do I find the perimeter of a rectangle that's in a circle when all I know is the diameter (6 inches) and the area of the square (15).
I've been at this for a week please help
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After reading your post, I have a question:
does the problem talks about a RECTANGLE or about a SQUARE ?
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When formulated in a correct way, this problem has a nice, short, simple and elegant " exact " solution.
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| Find the perimeter of a rectangle inscribed in a circle of the |
| diameter 6 inches, if the area of the rectangle is 15 sq. inches. |
+-----------------------------------------------------------------------+
Let x and y be the dimensions of the rectangle.
Then you have these two equations
x^2 + y^2 = 36 (1) (The Pythagoras, applied to the legs and the hypotenuse.
which is the diameter of the circle)
xy = 15 (2) (area equation).
Multiply equation (2) by 2 (both sides) and add it to equation (1). You will get
x^2 + 2xy + y^2 = 36 + 2*15,
or
(x + y)^2 = 66, which implies x + y = . (3)
The perimeter of the rectangle is 2x + 2y = ( from equation (3) ) = = 16.248 inches (approximately).
Thus you have BOTH "exact" solution and approximate (rounded) numerical value.
A square of area 15 can't be inscribed in a circle of diameter 6; so I will assume that the area of 15 is of a rectangle -- as suggested in the first phrase of your post -- instead of the area of a square.
Let x be one dimension of the rectangle
Then 15/x is the other dimension
The diameter of the circle is the diagonal of the rectangle:
The equation does not factor over the integers, so use a numerical method or a graphing calculator to find the possible values of x. I leave that to you.
Note the equation is an even function. It has four zeros -- two positive values and the opposites of those two.
Obviously the negative solutions make no sense in the problem.
The two positive solutions are the two dimensions of the rectangle.
