In the following figure, the diameter of circle is 3cm, AB and MN are two diameters such the 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:-
2√2−1
2√2−12
3√2−12
2√2−13
A
2√2−1
B
2√2−12
C
3√2−12
D
2√2−13
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Solution
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AB=MN=3cm
AEEB=NLLM=12
AB=3cm
⇒AE+EB
⇒3
AE+2AE=3
AE=1cm
NL=1cm
OA=AB2=1.5cm
ON=MN2=1.5cm
OE=OA−AE=1.5−1=0.5cm
OL=ON−LN=0.5cm
Quatrilateral EOLG is square
HL=EO=0.5cm
Join OD
ΔOLD,
OD2=OL2+DL2
DL2=(1.5)2−(0.5)2
DL=√2cm
DH=DL−HL−√2−12=2√2−12.
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