GMAT Club World Cup 2022 (DAY 9): Two circles are inscribed in a recta

Question

https://gmatclub.com/forum/download/file.php?id=67627 Two circles are inscribed in a rectangle, as shown above. What is the distance between the centers of these circles? A. 2 B. \sqrt{5} C. 3 D. \sqrt{10} E. 2\sqrt{5} Untitled.png

ThatDudeKnows · Answer

The diameter of each circle is 2. If the two circles were side-by-side and tangential, their centers would be separated by 2r = d= 2. But they are offset a bit vertically, so their centers are farther apart than 2. A is out. Do we want something a little more than 2 or a lot more than 2? Just a little. Certainly not more than 1.5x. C, D, and E are out.

Answer choice B.

Archit3110 · Answer

the value of digonal is 3^2+4^2 ; √25 ; 5
area of the half of the rectangle 3*4/2 ; 6
let xy be the distance of two center circle
xy = √( 3-2r)^2 + ( 4-2r)^2 
we can say three ∆ area 
3r/2 + 4r/2 + 5r/2 
12r/2 ; 6r
6r=6
r=1
substitute r in √( 3-2r)^2 + ( 4-2r)^2 
xy = √(1+4) ; √5
 OPTION B  \sqrt{5}

Berdiyor · Answer

Hello everyone,

The correct answer is B.
The detailed explanation of the solution is in the pic. attached

Iotaa · Answer

Refering to the picture, 3R (in red) = 3; R=1

Also, 4-2R=2.

Distance between the center will be 1ˆ2+2ˆ2= Dˆ2,

So D = \sqrt{5}

Option B is the correct answer

gmatophobia · Answer

In the figure below we extend the circle's radius to meet the sides and construct a right angled triangle as shown.

Attachment:

Screenshot 2022-07-21 214307.png [ 93.97 KiB | Viewed 3535 times ]

In \(\triangle ABC\) = \(\sqrt{(3-2r)^2 + (4-2r)^2}\) ----------- (1)

Now lets construct three triangles as shown below -

Attachment:

Screenshot 2022-07-21 214530.png [ 80.69 KiB | Viewed 3530 times ]

PM = 5 (3 - 4 -5 triplet)

Area of \(\triangle MPN\) = Area( \(\triangle MOP\) ) + Area( \(\triangle PON\) ) + Area( \(\triangle MON\) )

\(\frac{(4*3)}{2}\) = \(\frac{1}{2} * 4 * r +\frac{1}{2} * 3 * r + \frac{1}{2} * 5 * r\) = 6r

r = 1 -------------------- (2)

Substituting 2 in 1

AB =\( \sqrt{1^2 + 2^2}\) = \(\sqrt{5}\)

IMO B

soudipsengupta · Answer

Please refer to the attachment for the diagram and solution

bumpbot · Answer

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GMAT Club World Cup 2022 (DAY 9): Two circles are inscribed in a recta

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GMAT Club World Cup 2022 (DAY 9): Two circles are inscribed in a recta

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