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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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1.44 Cms
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OM=1.44cm
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1.45cm (using similar traingle properties)
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1.44
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The answer will be 1.4
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