ABCD वर्गामध्ये, कर्ण AC आणि BD O येथे छेदतात. ∠CAB चे कोन दुभाजक BD आणि BC ला अनुक्रमे F आणि G येथे भेटतात. OF ∶ CG च्या समान किती आहे?

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SSC CGL Tier 2 Quant Previous Paper 2 (Held On: 3 Feb 2022)
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  1. 1 ∶ 2
  2. 1 ∶ 3
  3. 1 ∶ \(\sqrt{2}\)
  4. 1 ∶ \(\sqrt{3}\)

Answer (Detailed Solution Below)

Option 1 : 1 ∶ 2
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Detailed Solution

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दिले:

ABCD एक चौरस आहे

कर्ण O येथे छेदतो

∠CAB चा कोन दुभाजक BD आणि BC ला F आणि G वर भेटतो

गणना:

F2 Savita SSC 5-5-22 D3

बाजू AB x असू द्या

तर, कर्ण AC = √2.x

⇒ OA = OB = OC = OD = AC/2 = x/√2 ----(i)

आता, ΔAOB मध्ये आणि कोन दुभाजक प्रमेयातून

AB : AO = BF : OF

⇒ x : (x/√2) = BF/OF

⇒ √2 : 1 = BF/OF

BF = √2y, OF = y ----(ii) समजा

पासून, BF + OF = OB = x/√2 [पासून (i)]

⇒ √2y + y = x/√2

\(y = \frac{x}{\sqrt2 . (\sqrt2 + 1)}\)

\(OF = \frac{x}{\sqrt2 . (\sqrt2 + 1)}\) ----(iii) [पासून (ii)]

आता, ΔABC मध्ये आणि कोन दुभाजक प्रमेयातून

AB : AC = BG : GC

⇒ x : √2 x = BG : GC

⇒ 1 : √2 = BG : GC

चला BG = z, GC = √2z ----(iv)

पासून, BG + GC = BC = x

z + √2z = x

\(z = \frac{x}{(\sqrt2 + 1)}\)

\(\sqrt2.z = \frac{\sqrt2x}{(\sqrt2 + 1)}\)

\(CG = \frac{\sqrt2x}{(\sqrt2 + 1)}\) ----(v )

आता (iii) आणि (v) वरून, आपल्याला मिळेल

OF : CG = \( \frac{x}{\sqrt2 . (\sqrt2 + 1)} : \frac{\sqrt2x}{(\sqrt2 + 1)}\)

OF : CG = \( \frac{1}{\sqrt2} : \frac{\sqrt2}{1}\)

⇒ OF : CG = 1 : 2

∴ आवश्यक गुणोत्तर 1 : 2 आहे.

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