Question : In a square ABCD, diagonals AC and BD intersect at O. The angle bisector of
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution :
Given: ABCD is a square. Diagonal intersects at O. The angle bisector of
Let's denote the side AB as \(x\). So, the diagonal AC = \(\sqrt{2}x\).
Now, in
We have \(\frac{AB}{AO} = \frac{BF}{OF}\)
\(⇒\frac{x}{\frac{x}{\sqrt{2}}} = \frac{BF}{OF}\)
Let's denote BF as \(\sqrt{2}y\) and OF as \(y\). From equation (i),
We know that \(BF + OF = OB = \frac{x}{\sqrt{2}}\)
\(⇒\sqrt{2}y + y = \frac{x}{\sqrt{2}}\)
\(⇒y = \frac{x}{\sqrt{2} + 2}\)
\(⇒OF = \frac{x}{\sqrt{2} + 2}\) ___ (iii)
In
We have \(\frac{AB}{AC} = \frac{BG}{GC}\)
\(⇒\frac{x}{\sqrt{2}x} = \frac{BG}{GC}\)
Let's denote BG as \(z\) and GC as \(\sqrt{2}z\).___ (iv)
Since \(BG + GC = BC = x\)
We have \(z + \sqrt{2}z = x\)
\(⇒z = \frac{x}{\sqrt{2} + 1}\)
\(⇒CG = \frac{2x}{\sqrt{2} + 1}\)___ (v)
Now from equations (iii) and (v),
We get \(OF : CG = \frac{x}{\sqrt{2} + 2} : \frac{2x}{\sqrt{2} + 1}\)
\(OF : CG = 1 : 2\).
Hence, the correct answer is
SSC CGL Complete Guide
Candidates can download this ebook to know all about SSC CGL.
