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figjam786
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If a square and triangle with sides measuring at 1 unit, have circles
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17 Jul 2022, 04:48
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Difficulty:

55% (hard)

Question Stats:

52% (03:18) correct 48% (02:02) wrong based on 25 sessions

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If a square and triangle with sides measuring at \(1 unit\), have circles inscribed. Find the distance \(X\) between the centers of these 2 circles A & B.

A. \({8- \sqrt{3}}\)

B. \(\sqrt{\frac{8-\sqrt{3}}{3}}\)

C. \(\sqrt{\frac{8-\sqrt{3}}{6}}\)

D. \(\frac{8- \sqrt{3}}{3}\)

E. \(\frac{8- \sqrt{3}}{6}\)
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AmanAngrish
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Schools: HBS '26 Stanford '26 Wharton '26 Kellogg '26 Sloan '26 Haas '26 Booth '26
GMAT 1: 600 Q47 V25
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If a square and triangle with sides measuring at 1 unit, have circles
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17 Jul 2022, 06:00
Unable to attach file in the reply section, however posting the approach to solve the problem
Main aim is to find the radius of both the circles(r1 and r2) so as to find the difference between the two radii(rdiff = r2-r1)(r2 being radius of circle inside square)
For r1:
Area of Big triangle = 3 x(Area of smaller traingles)
= 3*(1/2*r1(height)*1(base)
Therefore, r1 = 1/(2Sqrt(3))
Similary r2 = 1/2
rdiff = r2-r1
= (sqrt(3)-1)/(2*sqrt(3))

Next step would be to implement pythagoras theorem with distance bewteen traingle and square centres(i.e 1) as side A, diff between two radii and side B
Therefore
C^2 = A^2+B^2
= (1)^2 + (rdiff)^2
Upon solving we will get answer as option C.
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ThatDudeKnows
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If a square and triangle with sides measuring at 1 unit, have circles
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 Updated on: 22 Jul 2022, 13:51
Originally posted by ThatDudeKnows on 17 Jul 2022, 15:25.
Last edited by ThatDudeKnows on 22 Jul 2022, 13:51, edited 1 time in total.
figjam786
If a square and triangle with sides measuring at \(1 unit\), have circles inscribed. Find the distance \(X\) between the centers of these 2 circles A & B.

A. \({8- \sqrt{3}}\)

B. \(\sqrt{\frac{8-\sqrt{3}}{3}}\)

C. \(\sqrt{\frac{8-\sqrt{3}}{6}}\)

D. \(\frac{8- \sqrt{3}}{3}\)

E. \(\frac{8- \sqrt{3}}{6}\)
Show more

If you have NOOOOO idea how to do geometry, you still have at least a 50/50 and can probably get to something like 80/20.

Look at the thing. The horizontal distance between A and B is 1. The vertical distance is really small. We want an answer choice that's just barely more than 1. A, B, and D aren't close. We are down to two. If you don't have a clue what to do next, congrats on having a 50/50, since that's way better than a blind guess!!

If you use a little logic, you can see that we're going to need to apply the distance formula (or Pythagoras...same thing), and that's going to mean a square root. Depending on what question number you're on, how much time you have left, your overall pacing strategy, and how quickly you think you can work through the geometry (without making a careless mistake!), you could very well feel comfortable with C. You've done no math and have something like an 80/20!!

If you've got the geometry chops to actually work through the math, we can get that to 100/0.
Treat the bottom left of the diagram as the origin in a coordinate plane, so point B lies at (1.5, 0.5).
The triangle O-A-centerofbaseoftriangle is a 30-60-90. The longer leg is 1/2. So the shorter leg is (1/2)/\(\sqrt{3}\) = \(\sqrt{3}/6\)
Point A lies at (0.5, \(\frac{\sqrt{3}}{6}\)).
Making a right triangle between A and B, the horizontal leg is 1 and the vertical leg is \(\frac{1}{2}-\frac{\sqrt{3}}{6}=\frac{3-\sqrt{3}}{6}\).
\(d^2=1^2+(\frac{3-\sqrt{3}}{6})^2\)
\(d^2=1+(\frac{9-6\sqrt{3}+3}{36})\)
\(d^2=\frac{36}{36}+(\frac{12-6\sqrt{3}}{36})\)
\(d^2=\frac{48-6\sqrt{3}}{36}\)
\(d^2=\frac{8-\sqrt{3}}{6}\)
Take the square root of both sides.

Answer choice C.
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figjam786
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If a square and triangle with sides measuring at 1 unit, have circles
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22 Jul 2022, 10:43
ThatDudeKnows
figjam786
If a square and triangle with sides measuring at \(1 unit\), have circles inscribed. Find the distance \(X\) between the centers of these 2 circles A & B.

A. \({8- \sqrt{3}}\)

B. \(\sqrt{\frac{8-\sqrt{3}}{3}}\)

C. \(\sqrt{\frac{8-\sqrt{3}}{6}}\)

D. \(\frac{8- \sqrt{3}}{3}\)

E. \(\frac{8- \sqrt{3}}{6}\)

If you have NOOOOO idea how to do geometry, you still have at least a 50/50 and can probably get to something like 80/20.

Look at the thing. The horizontal distance between A and B is 1. The vertical distance is really small. We want an answer choice that's just barely more than 1. A, B, and D aren't close. We are down to two. If you don't have a clue what to do next, congrats on having a 50/50, since that's way better than a blind guess!!

If you use a little logic, you can see that we're going to need to apply the distance formula (or Pythagoras...same thing), and that's going to mean a square root. Depending on what question number you're on, how much time you have left, your overall pacing strategy, and how quickly you think you can work through the geometry (without making a careless mistake!), you could very well feel comfortable with C. You've done no math and have something like an 80/20!!

If you've got the geometry chops to actually work through the math, we can get that to 100/0.
Treat the bottom left of the diagram as the origin in a coordinate plane, so point B lies at (1.5, 0.5).
The triangle O-A-centerofbaseoftriangle is a 30-60-90. The longer leg is 1/2. So the shorter leg is (1/2)/\(\sqrt{3}\) = \(\sqrt{3}/6\)
Point A lies at (0.5, \(\frac{\sqrt{3}}{6}\)).
Making a right triangle between A and B, the horizontal leg is 1 and the vertical leg is \(\frac{1}{2}-\frac{\sqrt{3}}{6}=\frac{3-\sqrt{3}}{6}\).
\(d^2=1^2+(\frac{3-\sqrt{3}}{6})^2\)
\(d^2=1+(\frac{9-6\sqrt{3}+3}{36})\)
\(d^2=\frac{36}{36}+(\frac{12-6\sqrt{3}}{36})\)
\(d^2=\frac{48-6\sqrt{3}}{36}\)
\(d^2=\frac{8-\sqrt{3}}{6}\)
Take the square root of both sides.

Answer choice D.
Show more


I hope you meant answer choice C.

Posted from my mobile device
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ThatDudeKnows
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If a square and triangle with sides measuring at 1 unit, have circles
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22 Jul 2022, 13:51
figjam786
ThatDudeKnows
figjam786
If a square and triangle with sides measuring at \(1 unit\), have circles inscribed. Find the distance \(X\) between the centers of these 2 circles A & B.

A. \({8- \sqrt{3}}\)

B. \(\sqrt{\frac{8-\sqrt{3}}{3}}\)

C. \(\sqrt{\frac{8-\sqrt{3}}{6}}\)

D. \(\frac{8- \sqrt{3}}{3}\)

E. \(\frac{8- \sqrt{3}}{6}\)

If you have NOOOOO idea how to do geometry, you still have at least a 50/50 and can probably get to something like 80/20.

Look at the thing. The horizontal distance between A and B is 1. The vertical distance is really small. We want an answer choice that's just barely more than 1. A, B, and D aren't close. We are down to two. If you don't have a clue what to do next, congrats on having a 50/50, since that's way better than a blind guess!!

If you use a little logic, you can see that we're going to need to apply the distance formula (or Pythagoras...same thing), and that's going to mean a square root. Depending on what question number you're on, how much time you have left, your overall pacing strategy, and how quickly you think you can work through the geometry (without making a careless mistake!), you could very well feel comfortable with C. You've done no math and have something like an 80/20!!

If you've got the geometry chops to actually work through the math, we can get that to 100/0.
Treat the bottom left of the diagram as the origin in a coordinate plane, so point B lies at (1.5, 0.5).
The triangle O-A-centerofbaseoftriangle is a 30-60-90. The longer leg is 1/2. So the shorter leg is (1/2)/\(\sqrt{3}\) = \(\sqrt{3}/6\)
Point A lies at (0.5, \(\frac{\sqrt{3}}{6}\)).
Making a right triangle between A and B, the horizontal leg is 1 and the vertical leg is \(\frac{1}{2}-\frac{\sqrt{3}}{6}=\frac{3-\sqrt{3}}{6}\).
\(d^2=1^2+(\frac{3-\sqrt{3}}{6})^2\)
\(d^2=1+(\frac{9-6\sqrt{3}+3}{36})\)
\(d^2=\frac{36}{36}+(\frac{12-6\sqrt{3}}{36})\)
\(d^2=\frac{48-6\sqrt{3}}{36}\)
\(d^2=\frac{8-\sqrt{3}}{6}\)
Take the square root of both sides.

Answer choice D.


I hope you meant answer choice C.

Posted from my mobile device
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Oy, yeah...everything leading up to typing in the actual letter is correct...
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