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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth

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Math Expert V
Joined: 02 Sep 2009
Posts: 59039
In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 23% (03:46) correct 77% (02:43) wrong based on 63 sessions

### HideShow timer Statistics In the diagram, AB = 8 and AD = 9. Two circles are tangent to each other and also tangent to the sides of the rectangle, as shown. If the radius of the smaller circle is 2, what is the radius of the larger circle?

(A) $$\frac{5}{2}$$

(B) $$2\sqrt{2}$$

(C) $$3$$

(D) $$2\sqrt{3}$$

(E) $$\frac{7}{2}$$

Attachment: geo_q8.jpg [ 5.05 KiB | Viewed 960 times ]

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Senior Manager  V
Joined: 22 Feb 2018
Posts: 415
Re: In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth  [#permalink]

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2
3
OA: C
Attachment: circle.PNG [ 39.79 KiB | Viewed 864 times ]

Let the radius of big circle be R , Then $$AI =BG=R$$
Small circle radius $$=O_{S}E=O_{S}F=JG=HB=2cm$$
$$AB=8=AI+IH+HB=R+IH+2$$
$$IH=8-2-R =6-R$$

$$AD=BC= 9$$
$$BC=BG+GF+FC=R+GF+2$$
$$GF =9-2-R =7-R$$

$$O_{S}O_{B}=R+2$$

In $$\triangle O_{S}O_{B}J$$,

$$(O_{S}O_{B})^2 = (O_{S}J)^2 +(O_{B}J)^2$$
$$O_{S}J = GF$$
$$O_{B}J = IH$$

$$(R+2)^2=(6-R)^2+(7-R)^2$$
$$R^2+4+4R = 36 +R^2 -12R +49 +R^2 -14R$$
$$R^2+4R+4 = 2R^2 -26R +85$$
$$R^2 -30R +81=0$$
Solving for R , we get R=3cm or 27 cm( neglecting 27 cm as it is not possible to have circle with R=27 inscribed inside 8*9rectangle)
Final $$R=3 cm$$
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##### General Discussion
Intern  B
Joined: 13 Jan 2016
Posts: 8
Re: In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth  [#permalink]

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Princ wrote:
OA: C
Attachment:
circle.PNG

Let the radius of big circle be R , Then $$AI =BG=R$$
Small circle radius $$=O_{S}E=O_{S}F=JG=HB=2cm$$
$$AB=8=AI+IH+HB=R+IH+2$$
$$IH=8-2-R =6-R$$

$$AD=BC= 9$$
$$BC=BG+GF+FC=R+GF+2$$
$$GF =9-2-R =7-R$$

$$O_{S}O_{B}=R+2$$

In $$\triangle O_{S}O_{B}J$$,

$$(O_{S}O_{B})^2 = (O_{S}J)^2 +(O_{B}J)^2$$
$$O_{S}J = GF$$
$$O_{B}J = IH$$

$$(R+2)^2=(6-R)^2+(7-R)^2$$
$$R^2+4+4R = 36 +R^2 -12R +49 +R^2 -14R$$
$$R^2+4R+4 = 2R^2 -26R +85$$
$$R^2 -30R +81=0$$
Solving for R , we get R=3cm or 27 cm( neglecting 27 cm as it is not possible to have circle with R=27 inscribed inside 8*9rectangle)
Final $$R=3 cm$$

@bunnel Is this explanation even correct i mean is this method correct ?
Manager  G
Joined: 14 Jun 2018
Posts: 212
In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth  [#permalink]

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1
Diagonal of the rectangle = √($$8^2 + 9^2$$) = √145
Let the radius of larger circle be x.

x√2 + x + 2√2 + 2 = √145 (diagonal of larger square + radius of larger circle + diagonal of smaller square + radius of smaller circle)

2.4 x = 12.1-4.82
x = 7.2/2.414 = 2..98.. = 3

Ans C
Attachments Capture.PNG [ 14.73 KiB | Viewed 463 times ] In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth   [#permalink] 06 Oct 2018, 12:08
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# In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth  