Hi dferm,
To solve this question, remember the rule for 45/45/90 degree triangles: the hypothenuse of these triangles are always 'a*sqrt2', assuming the other two sides are 'a'.
First, from the question, we know that each of the sides of the square S is 6 (perimeter is 24, 24/4 = 6). To get the diameter of the circle T, we need to find the hypothenuse of the triangles (i.e., the diagonal of the square) that we get when we cut the square in half, which is 6*sqrt2, according to our handy formula above. The radius is then 3*sqrt2.
Finally, the circumference of the circle is 2*pi*r, and this comes out to 6*sqrt2*pi.
Hope this helps!
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GMAT Prep(Square Inscribed) Pract2
Source: Beat The GMAT — Problem Solving |
Hi dferm,
It might help if you draw a picture and start filling in what I wrote above. I know it's a little weird that I'm telling you to work with triangles when the problem gives you a circle and a square. But, if you draw a picture, you'll see that when you cut the square in half diagonally you get two triangles, the hypotenuses of which are the diameter of the circle.
Does this clear it up?
It might help if you draw a picture and start filling in what I wrote above. I know it's a little weird that I'm telling you to work with triangles when the problem gives you a circle and a square. But, if you draw a picture, you'll see that when you cut the square in half diagonally you get two triangles, the hypotenuses of which are the diameter of the circle.
Does this clear it up?
Jim S. | GMAT Instructor | Veritas Prep
side of square=24/4=6
diagonal= 6*sqrt2 (as it is 45,90,45 triangle)
so, dia of circle=6*sqrt2
perimeter= pie*dia= 6*sqrt2*pie
diagonal= 6*sqrt2 (as it is 45,90,45 triangle)
so, dia of circle=6*sqrt2
perimeter= pie*dia= 6*sqrt2*pie
Why the eyeroll? Draw a picture. The diagonal of the square is equal to the diameter of the circle. 2r = d. Therefore C = 2rπ = dπ.vladmire wrote:I wanted to make sure I understand this the diagonal of the square is equal to the circles radius right/ Then d(pi) is equal to the circumference of the circle.
Before I under the impression that 2*pi*r = circumference. :roll:
Guys,
I clearly understand the explanation above. But what is the reason to equate diagnoal of the square to the diameter of the circle?
Vidhya
I clearly understand the explanation above. But what is the reason to equate diagnoal of the square to the diameter of the circle?
Vidhya
Draw a figure to see the relation...
The diagonal of the square = Diameter of the circle..
You could also do it using 45-45-90 triangles.
The diagonal of the square = Diameter of the circle..
You could also do it using 45-45-90 triangles.
vidhya16 wrote:Guys,
I clearly understand the explanation above. But what is the reason to equate diagnoal of the square to the diameter of the circle?
Vidhya
Square S is inscribed in circle T. If the perimeter of S is 24, what is the circumference of T?
A. 6Ï€
B. 12Ï€
C. 3√2π
D. 6√2π
E. 12√2π
Draw the figure:vidhya16 wrote:Guys,
I clearly understand the explanation above. But what is the reason to equate diagnoal of the square to the diameter of the circle?
Vidhya

An inscribed angle is formed by 2 chords.
Angle BAD is an inscribed angle.
Since ABCD is a square, angle BAD = 90.
An inscribed angle of 90 degrees intercepts the DIAMETER of the circle.
For this reason, when a square is inscribed in a circle, the DIAGONAL of the square is also the DIAMETER of the circle.
The diagonal of a square forms a 45-45-90 triangle.
In a 45-45-90 triangle, the sides are proportioned s:s:s√2.
Since the perimeter of the square = 24, s=6, implying that BD = 6√2.
Thus, the circumference of the circle = πd = 6√2π.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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