GRE Math : Diameter
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Example Questions
Example Question #1 : Diameter
The formula to find the radius of the largest circle that can fit in an equilateral triangle is 
What is the largest diameter of a circle that can fit inside an equilateral triangle with a perimeter of 
The diameter is
To solve for the largest diameter multiply each side by 2.
The resulting formula for diamenter is
Substitute in 5 for S and solve. Diameter = 
Example Question #2 : Diameter
Quantity A: The diameter of a circle with area of
Quantity B: The diameter of a circle with circumference of
Which of the following is true?
Quantity B is larger.
The relationship of the quantities cannot be determined.
Quantity A is larger.
Both quantities are equal.
Quantity B is larger.
Consider each quantity separately.
Quantity A
Recall that the area of a circle is defined as:
We know that the area is 
Divide both sides by 
Therefore, 
Quantity B
This is very easy. Recall that:
Therefore, if 
Example Question #1 : How To Find The Length Of The Diameter
Quantity A: The diameter of a circle with area of
Quantity B: The diameter of a circle with circumference of
Which of the following is true?
The relationship between the quantities cannot be determined.
The two quantities are equal.
Quantity B is larger.
Quantity A is larger.
Quantity B is larger.
Consider each quantity separately.
Quantity A
Recall that the area of a circle is defined as:
We know that the area is 
Divide both sides by 
Therefore, 
Quantity B
This is very easy. Recall that:
Therefore, if 
Now, since your calculator will not have a square root button on it, we need to estimate for Quantity A. We know that 
Example Question #4 : Diameter
A circle with an area of 
From the ratio given 
or
The largest sector thus has an area equal to
Example Question #5 : Diameter
A rectangle is inscribed inside of a circle such that every corner touches the edge of the circle. If the area of the rectangle is 
The answer cannot be determined.
To find the area of the circle, it is important to know either its diameter or radius. For the geometry described in this problem, this is the same as the diagonal of the rectangle.
However, to find the diagonal of the rectangle, the sides must first be known. They can be found, since the perimeter and area are given:
This system of equation can be solved by substitution:
Followed by:
Note that this gives two possible values for 
Knowing these two values, the diagonal can be found; it is the hypotenuse of a right triangle formed by these two lengths:
Since the diagonal is also the diameter, the area of the circle is given by:
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