The angles of a quadrilateral ABCD taken in an order are in the ratio \(3:7:6:4\) . Then \(ABCD\) is a ( )
A. kite
B. parallelogram
C. rhombus
D. trapezium
Answer
D
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Solution
Let the angles of quadrilateral \(ABCD\) be \(\angle A=3x,\angle B=7x,\angle C=6x,\angle D=4x\) We know that sum of all angles of a quadrilateral \(=360^{\circ}\) . \(\Rightarrow 3x+7x+6x+4x=360^{\circ}\). \(\Rightarrow 20x=360^{\circ}\) \(x= \frac{360^{\circ}}{20} = 18^{\circ}\). Hence, \(\angle A=3x=3\times 18^{\circ}=54^{\circ}\), \(\angle B = 7\times 18^{\circ}= 126^{\circ}\), \(\angle C=6x=6\times 18^{\circ}= 108^{\circ}\), \(\angle D = 4x=4\times 18^{\circ}=72^{\circ}\). \(\angle A + \angle B = 54^{\circ} +126^{\circ} =180^{\circ}\) (Satisfying the corresponding angles)...(\(1\)) \(\angle C +\angle D = 108^{\circ}+72^{\circ}=180^{\circ}\) (Satisfying the corresponding angles)...(\(2\)) From \((1)\) and \((2)\) Since the corresponding angles property is satisfies. Hence, \(AB\parallel CD\). Hence, the given figure is a trapezium.