Length of segment in a circle geometry construction

CAT 2005 Slot 1 · QA · Hard · Geometry

All CAT 2005 Slot 1 questionsMore Geometry questionsAll CAT QA

In the given figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB=1:2AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM=1:2NL:LM = 1:2. The length of DH in cm is

  1. A.

    22−12\sqrt{2} - 1

  2. B.

    22−12\frac{2\sqrt{2} - 1}{2}

  3. C.

    32−12\frac{3\sqrt{2} - 1}{2}

  4. D.

    22−13\frac{2\sqrt{2} - 1}{3}

Answer

B

Explanation

Radius R=1.5=32R = 1.5 = \frac{3}{2} cm. Since AE:EB=1:2AE : EB = 1 : 2 and AB=3AB = 3, AE=1AE = 1 cm, so OE=0.5=12OE = 0.5 = \frac{1}{2} cm. Since NL:LM=1:2NL : LM = 1 : 2, NL=1NL = 1 cm, so OL=0.5=12OL = 0.5 = \frac{1}{2} cm. Note that HL=OE=12HL = OE = \frac{1}{2} cm. In ΔDOL\Delta DOL, DO=R=32DO = R = \frac{3}{2} cm, OL=12OL = \frac{1}{2} cm. DL=DO2−OL2=94−14=2DL = \sqrt{DO^2 - OL^2} = \sqrt{\frac{9}{4} - \frac{1}{4}} = \sqrt{2}. Since DL=DH+HL  ⟹  2=DH+12  ⟹  DH=2−12=22−12DL = DH + HL \implies \sqrt{2} = DH + \frac{1}{2} \implies DH = \sqrt{2} - \frac{1}{2} = \frac{2\sqrt{2} - 1}{2}.

Practise this under exam conditions

Sign in to solve it with a live timer, the on-screen CAT calculator, and streak and accuracy tracking across every question you attempt.

Solve in the workspace