Please submit your concern

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

Explanation:

AO = OD = 1.5 cm

 AE + EB = 3 cm and AE : EB = 1 : 2

∴ AE = 1 cm and EB = 2 cm

OE = AO – AE = 1.5 – 1 = 0.5 cm

Similarly, NL = 1 cm, LM = 2 cm, and OL = 0.5 cm

OEHL is a square as all its angles are right angles and OE = OL

∴ EH = HL = 0.5 cm

In ∆ODL, OD² = OL² + DL²

1.5² = 0.5² + (0.5 + DH)²

2.25 = 0.25 + 0.25 + DH + DH²

DH² + DH – 1.75 = 0

DH = $\frac{-1\pm \sqrt{1-4(-1.75)}}{2}$ = $\frac{2\sqrt{2}-1}{2}$ (DH > 0)

Hence, option (b).