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25 Jul 2018, 00:35
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (03:04) correct
68%
(02:53)
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 3064 times ]
25 Jul 2018, 03:26
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OA: C
circle.PNG [ 39.79 KiB | Viewed 2813 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 \(R=3 cm\)
Attachment:
circle.PNG [ 39.79 KiB | Viewed 2813 times ]
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 \(R=3 cm\)
General Discussion
06 Oct 2018, 07:47
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Princ
OA: C
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 \(R=3 cm\)
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 ?
