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 $\mathrm{AB}$.

Answer 1

Finding the length of $\mathrm{AB}$:

Given:

$\mathrm{ABCD}$ is a trapezium with$\mathrm{AB}\left|\right|\mathrm{CD}.$

⇒$\angle \mathrm{ABC}+\angle \mathrm{ADC}=180°$

From the given details we know that

$\angle \mathrm{ADC}=2(\angle \mathrm{ABC})$

⇒ $\angle \mathrm{ABC}+2(\angle \mathrm{ABC})=180º$

$\Rightarrow 3(\angle \mathrm{ABC})=180º$

$\Rightarrow \angle \mathrm{ABC}=\frac{180º}{3}=60º$

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$\angle \mathrm{DAE}=60°.$

$\cos 60º=\frac{1}{2}=\frac{\mathrm{AE}}{\mathrm{AD}}$

Hence,

$AE=\frac{AD}{2}=\frac{a}{2}$

$\mathrm{EF}$is equal to$\mathrm{CD}=\mathrm{b}$

$\mathrm{AB}=\mathrm{AE}+\mathrm{EF}+\mathrm{FB}$

$\begin{array}{rcl}& =& \frac{\mathrm{a}}{2}+\mathrm{b}+\frac{\mathrm{a}}{2}\\ & =& (\frac{\mathrm{a}}{2}+\frac{\mathrm{a}}{2})+\mathrm{b}\\ & =& \mathrm{a}+\mathrm{b}\end{array}$

Therefore the length of $\mathrm{AB}$ is $\begin{array}{rcl}& & \mathrm{a}+\mathrm{b}\end{array}$