Question

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 \( AE : EB = 1 : 2 \), 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. \( \frac{2\sqrt{2}-1}{2} \)
• C. \( \frac{3\sqrt{2}-1}{2} \)
• D. \( \frac{2\sqrt{2}-1}{3} \)

Hint:

In geometry problems involving ratios and right-angle triangles, always apply the Pythagorean theorem and the properties of similar triangles to break down the problem into smaller, solvable parts.

Answer 1

The Correct Option is B

Step 1: Understand the geometric setup
The figure is a combination of geometric elements including a circle with two perpendicular diameters, AB and MN, intersecting at the center. Points \( CG \) and \( DF \) are drawn perpendicular to the diameters.

Step 2: Use given ratio information
We are given that the ratio \( AE : EB = 1 : 2 \), which implies that the line segment \( AB \) is divided in a ratio of 1:2 at point \( E \). Similarly, the ratio \( NL : LM = 1 : 2 \) divides the segment \( MN \).

Step 3: Apply geometric principles
First, note that since \( AE : EB = 1 : 2 \), we can calculate that: \[ AE + EB = AB = 3 \quad \text{(as \( AB = 3 \))} \] \[ AE = 1 \quad \text{and} \quad EB = 2 \] Thus, \( AB = 3 \) and \( AE = 1 \).
Next, we apply this ratio in the context of triangle \( OLD \), where: \[ % Option (OD)^2 = (OL)^2 + (DL)^2 \] We use this formula to calculate \( DL \). \[ DL = \sqrt{2} \quad \text{from the relation for triangle \( OLD \)} \]

Step 4: Solve for DH using the Pythagorean theorem
Now we can solve for the length of \( DH \) based on the geometric properties of the figure: \[ DH = DL - HL = \sqrt{2} - 1 \] This simplifies to the required result: \[ DH = \frac{2\sqrt{2} - 1}{2} \] Final Answer: The correct answer is (b) \( \frac{2\sqrt{2}-1}{2} \).