Question

The measure of the angles of the quadrilateral ABCD are in the ratio 2 : 3 : 4 : 1 Determine the type of the quadrilateral

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{common}\mathrm{multiple}\mathrm{of}\mathrm{the}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{angles}\mathrm{be}x. \\ \mathrm{Let}: \\ \angle \mathrm{A}=\left(2x\right)° \\ \angle \mathrm{B}=\left(3x\right)° \\ \angle \mathrm{C}=\left(4x\right)° \\ \angle \mathrm{D}=\left(x\right)° \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{quadrilateral}\mathrm{is}360°. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ 2x+3x+4x+x=360 \\ \Rightarrow 10x=360 \\ \Rightarrow x=36 \\ \mathrm{Hence},\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}\mathrm{are}: \\ \angle \mathrm{A}=\left(2\times 36\right)°=72° \\ \angle \mathrm{B}=\left(3\times 36\right)°=108° \\ \angle \mathrm{C}=\left(4\times 36\right)°=144° \\ \angle \mathrm{D}=\left(1\times 36\right)°=36° \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \angle \mathrm{A}+\angle \mathrm{D}=72°+36°=108°\ne 180° \\ \angle \mathrm{B}+\angle \mathrm{C}=108°+144°=252°\ne 180° \\ \mathrm{That}\mathrm{means}\mathrm{AB}\nparallel \mathrm{CD}. \\ \mathrm{But}\angle \mathrm{A}+\angle \mathrm{B}=72°+108°=180°=180° \\ \angle \mathrm{C}+\angle \mathrm{D}=144°+36°=180°=180° \\ \mathrm{That}\mathrm{means}\mathrm{AD}\parallel \mathrm{BC} \\ \mathrm{Hence},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}\mathrm{the}\mathrm{given}\mathrm{quadrilateral}\mathrm{is}\mathrm{a}\mathrm{trapezium}. \end{gathered}$$