If angles A, B, C and D of the quadrilateral | vedclass

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(C) Let the angles $A, B, C$ and $D$ of the quadrilateral $ABCD$ be $3x, 7x, 6x$ and $4x$ respectively.Since the sum of the angles of a quadrilateral

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trapezium

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(C) Let the angles $A, B, C$ and $D$ of the quadrilateral $ABCD$ be $3x, 7x, 6x$ and $4x$ respectively.Since the sum of the angles of a quadrilateral is $360^{\circ}$,we have:$3x + 7x + 6x + 4x = 360^{\circ}$$20x = 360^{\circ}$$x = 18^{\circ}$Thus,the angles are:$A = 3 	imes 18^{\circ} = 54^{\circ}$$B = 7 	imes 18^{\circ} = 126^{\circ}$$C = 6 	imes 18^{\circ} = 108^{\circ}$$D = 4 	imes 18^{\circ} = 72^{\circ}$Now,consider the sum of consecutive angles $\angle C$ and $\angle D$:$\angle C + \angle D = 108^{\circ} + 72^{\circ} = 180^{\circ}$Since the sum of the interior angles on the same side of the transversal $CD$ is $180^{\circ}$,the sides $AD$ and $BC$ must be parallel $(AD \parallel BC)$.Since $ABCD$ is a quadrilateral with one pair of opposite sides parallel,it is a trapezium.

If angles $A, B, C$ and $D$ of the quadrilateral $ABCD$,taken in order,are in the ratio $3:7:6:4$,then $ABCD$ is a

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