The angles of a quadrilateral ABCD taken in an order are in the ratio \(3:7:6:4\) . Then \(ABCD\) is a （ ）
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The angles of a quadrilateral ABCD taken in an order are in the ratio \(3:7:6:4\) . Then \(ABCD\) is a （ ）
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.
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.
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