Question

In the following 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 = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is

• A. \(2\sqrt{2} - 1\)
• B. \((2\sqrt{2} - 1)/2\)
• C. \((3\sqrt{2} - 1)/2\)
• D. \((2\sqrt{2} - 1)/3\)

Hint:

Place the centre O at the origin. E and L are each half a centimetre from O, so H is the point (-0.5, -0.5). Then find where the horizontal line through H meets the circle.

Answer 1

The Correct Option is B

Step 1: Set up coordinates:
The radius is \(1.5\) cm. Take \(O = (0,0)\), AB along the x-axis (A on the left, B on the right) and MN along the y-axis (M at the top, N at the bottom).

Step 2: Locate E, L and H:
\(AB = 3\) and \(AE : EB = 1 : 2\), so \(AE = 1\). Then \(E\) is \(1.5 - 1 = 0.5\) to the left of \(O\), and the line CG is \(x = -0.5\). \(MN = 3\) and \(NL : LM = 1 : 2\), so \(NL = 1\), and \(L\) is \(0.5\) below \(O\); the line DF is \(y = -0.5\). Hence \(H = (-0.5, -0.5)\).

Step 3: Find D:
D lies on the circle \(x^2 + y^2 = \tfrac{9}{4}\) with \(y = -0.5\) on the left side: \(x^2 = \tfrac94 - \tfrac14 = 2\), so \(x = -\sqrt2\). Thus \(D = (-\sqrt2, -0.5)\).

Step 4: Find DH:
D and H have the same y-coordinate, so \(DH = -0.5 - (-\sqrt2) = \sqrt2 - \tfrac12 = \dfrac{2\sqrt2 - 1}{2}\) cm.

Final Answer:
The correct option is (2): \((2\sqrt{2} - 1)/2\)