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M is the midpoint of AB, so AM = MB.

MB = ½AB = ½ x 4cm = 2cm

If you managed to do the first part, you would know by now that the hexagon can be divided into 6 equilateral triangles. One of them is △OAB. Since AB is a side of the equilateral △ and AB = 4cm, OB = 4cm since all 3 sides of an equilateral triangle are equal in length.

△OAB is divided by line OM into two congruent right-angled triangles. △OMB is one of them.

Using Pythagoras' theorem, OM² + MB² = OB²

OM² = OB² - MB² = (4cm)² - (2cm)²

OM² = 16cm² - 4cm² = 12cm²

OM = √12 cm = √(4x3) cm = 2√3 cm

Now, △OMX is also a right-angled triangle.

So using Pythagoras' theorem again,

OM² + OX² = XM²

12cm² + OX² = (10cm)²

OX² = 100cm² - 12cm² = 88cm²

OX = √88 cm = √(4 x 22) cm = 2√22 cm

MB = ½AB = ½ x 4cm = 2cm

If you managed to do the first part, you would know by now that the hexagon can be divided into 6 equilateral triangles. One of them is △OAB. Since AB is a side of the equilateral △ and AB = 4cm, OB = 4cm since all 3 sides of an equilateral triangle are equal in length.

△OAB is divided by line OM into two congruent right-angled triangles. △OMB is one of them.

Using Pythagoras' theorem, OM² + MB² = OB²

OM² = OB² - MB² = (4cm)² - (2cm)²

OM² = 16cm² - 4cm² = 12cm²

OM = √12 cm = √(4x3) cm = 2√3 cm

Now, △OMX is also a right-angled triangle.

So using Pythagoras' theorem again,

OM² + OX² = XM²

12cm² + OX² = (10cm)²

OX² = 100cm² - 12cm² = 88cm²

OX = √88 cm = √(4 x 22) cm = 2√22 cm

∠MOB = 30° since OM bisects ∠AOB, which is 60°.

cos30° = OM/OB = OM/4cm

OM = 4(cos30°)cm = 4(√3 / 2)cm = 2√3cm