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ABCD is a cyclic trapezium with AB the diameter, AB ∥ CD, BC = 4 cm and AB = 10 cm, then find the length of side CD.
- A6.8 cm
- B6.4 cm
- C6.6 cm
- D6 cm
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CCommunity
Given : AB is a diameter
$ \Large \angle CAB=30 ^{\circ} $
As we know that
$ \Large \angle ACB=90 ^{\circ} $
$ \Large \therefore \angle ACB+ \angle CAB+ \angle CBA=180 ^{\circ} $
$ \Large \angle CBA=180 ^{\circ} -90 ^{\circ} -30 ^{\circ} $
$ \Large \angle CBA=60 ^{\circ} $
$\textbf{Note:}$ In a cyclic trapezium sum of opposite angle is $180 ^{\circ} $
$ \Large \therefore \angle D+ \angle B=180 ^{\circ} $
$ \Large \angle D=180 ^{\circ} -60 ^{\circ} $
$ \Large \angle D=120 ^{\circ} $
$ \Large \angle CAB=30 ^{\circ} $
As we know that
$ \Large \angle ACB=90 ^{\circ} $
$ \Large \therefore \angle ACB+ \angle CAB+ \angle CBA=180 ^{\circ} $
$ \Large \angle CBA=180 ^{\circ} -90 ^{\circ} -30 ^{\circ} $
$ \Large \angle CBA=60 ^{\circ} $
$\textbf{Note:}$ In a cyclic trapezium sum of opposite angle is $180 ^{\circ} $
$ \Large \therefore \angle D+ \angle B=180 ^{\circ} $
$ \Large \angle D=180 ^{\circ} -60 ^{\circ} $
$ \Large \angle D=120 ^{\circ} $