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DA,CBand OM are each perpendicular to line segment AB, where O is the point of intersection ofAC and DB. if AD=2.4cm, CB=3.6cm then OM=

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In a right triangle BAD, We have OM parallel to side DA of triangle BAD. So triangle BMO is similar to triangle BAD by AA similarity criteria. Thus corresponding sides of similar triangles are proportional. Thus, OM÷AD = BM÷AB, which implies, OM÷2.4 = BM÷AB ...........(1)...
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In a right triangle BAD,
We have OM parallel to side DA of triangle BAD. So triangle BMO is similar to triangle BAD by AA similarity criteria.
Thus corresponding sides of similar triangles are proportional.
Thus, OM÷AD = BM÷AB, which implies, OM÷2.4 = BM÷AB ...........(1)
In the same way, triangle ABC is similar to triangle AMO.
Thus, we have OM÷3.6 = AM÷AB .........(2)
(1)+(2) gives (OM÷2.4)+(OM÷3.6)=(BM÷AB)+(AM÷AB)
On simplifying this,
we get OM = (3.6 × 2.4) ÷ 6 = 1.44 cm

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A Mathematician With 2-3 Years of Experience teaching students & helping them with sums

1.44 Cms
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1.44 cm
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OM=1.44cm
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Maths and physics tutor since past years

1.45cm (using similar traingle properties)
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1.44
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Teaches since 2015 online and offline. Understanding and clarity of subject is my first priority.

The answer will be 1.4
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M.sc chemistry with 1year experience of teaching

1.44
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2.2
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PCMB 4 YR Experience of teaching

1.4cm
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