The diameter of the circle is 3 cm.AB and MN are two diameters such that MN is perpendicular to AB.In addition

Question

The diameter of the circle is 3 cm.AB(horizontally drawn) and MN(vertically drawn) 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.C,G,D and F are on the circumference of the circle.CG is drawn to the left of MN and DF is drawn below AB.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

Answer 1

I assume that E, L, and H are the points of intersection of AB and CG, MN and DF, and CG and DF, respectively. AB and MN are each of length 3cm so AE and NL are each 1cm and EB and ML are each 2cm. So, if O is the centre of the circle, then EO and OL are each 1/2cm. Since CG and MN are parallel and AB and DF are parallel it follows that EOLH is a square so LH is 1/2cm. Now DOL is a right-angled triangle with DO 1.5cm, and OL 0.5cm, so DL^2 +OL^2 = DO^2
i.e. DL^2 = (1.5)^2 - (0.5)^2 =2.25 - 0.25 = 2. Thus DL =sqrt(2) and so DH = DL - 1/2 =sqrt(2) - 1/2 =(2*sqrt(2) - 1)/2. hence the correct answer is (b).

Answer 2

where is H?

Answer 3

Where is H??????
What is DH??????