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Question

# O is the centre of the circle having radius 5 cm. AB and AC are two chords such that AB=AC=6 cm. If OA meets BC at M, then OM is equal to

A
3.6 cm
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B
1.4 cm
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C
2 cm
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D
3 cm
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Solution

## The correct option is B 1.4 cmM is the mid point of BCThus, BM=MC=x [Let]OA is the radius of the circleLet AM=yOM=OA−AMOM=5−y ----(i)Now, △ABM is a right angled triangle∴AB2=AM2+BM2=>62=y2+x2=>x2=36−y2 -----(ii)Again in △OMBOB2=OM2+BM2=>52=(5−y)2+x2 ....[Using (i)]=>25=25−10y+y2+36−y2 ....[Using (ii)]=>10y=36=>y=3.6Thus, OM=(5−3.6)=1.4cm

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