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# In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth

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

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25 Jul 2018, 00:35
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Difficulty:

95% (hard)

Question Stats:

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

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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 936 times ]

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

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25 Jul 2018, 03:26
2
3
OA: C
Attachment:

circle.PNG [ 39.79 KiB | Viewed 844 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
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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06 Oct 2018, 07:47
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
Joined: 14 Jun 2018
Posts: 219
In the diagram, AB = 8 and AD = 9. Two circles are tangent to each oth  [#permalink]

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06 Oct 2018, 12:08
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
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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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