Question

$\mathrm{ABCD}$is a trapezium in which $AB\mathit{}II\mathit{}CD$. If $\angle \mathrm{ADC}$ is equal to twice $\angle \mathrm{ABC}$, $\mathrm{AD}=\mathrm{a}\mathrm{cm}$ and $\mathrm{CD}=\mathrm{b}\mathrm{cm}$, then find the length of $\mathrm{AB}$.

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Solution

Finding the length of $\mathrm{AB}$:Given:$\mathrm{ABCD}$ is a trapezium with$\mathrm{AB}||\mathrm{CD}.$⇒$\angle \mathrm{ABC}+\angle \mathrm{ADC}=180°$ From the given details we know that$\angle \mathrm{ADC}=2\left(\angle \mathrm{ABC}\right)$⇒ $\angle \mathrm{ABC}+2\left(\angle \mathrm{ABC}\right)=180º$$⇒3\left(\angle \mathrm{ABC}\right)=180º$$⇒\angle \mathrm{ABC}=\frac{180º}{3}=60º$Let us draw perpendiculars from the vertices $\mathrm{D},\mathrm{C}$to $\mathrm{AB}$, namely $\mathrm{DE}$ and$\mathrm{CF}$From the figure, $\mathrm{DE}$is perpendicular to$\mathrm{AB}$, then triangle $\mathrm{ADE}$ is formed with$\angle \mathrm{DAE}=60°.$$\mathrm{cos}60º=\frac{1}{2}=\frac{\mathrm{AE}}{\mathrm{AD}}$Hence,$AE=\frac{AD}{2}=\frac{a}{2}$$\mathrm{EF}$is equal to$\mathrm{CD}=\mathrm{b}$$\mathrm{AB}=\mathrm{AE}+\mathrm{EF}+\mathrm{FB}$ $\begin{array}{rcl}& =& \frac{\mathrm{a}}{2}+\mathrm{b}+\frac{\mathrm{a}}{2}\\ & =& \left(\frac{\mathrm{a}}{2}+\frac{\mathrm{a}}{2}\right)+\mathrm{b}\\ & =& \mathrm{a}+\mathrm{b}\end{array}$Therefore the length of $\mathrm{AB}$ is $\begin{array}{rcl}& & \mathrm{a}+\mathrm{b}\end{array}$

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