Question

The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{ABCD}\mathrm{be}\mathrm{the}\mathrm{trapezium}\mathrm{in}\mathrm{which}\mathrm{AB}\parallel \mathrm{DC},\mathrm{AB}=25\mathrm{cm},\mathrm{CD}=11\mathrm{cm},\mathrm{AD}=13\mathrm{cm}\mathrm{and}\mathrm{BC}=15\mathrm{cm}. \\ \mathrm{Draw}\mathrm{CL}\perp \mathrm{AB}\mathrm{and}\mathrm{CM}\parallel \mathrm{DA}\mathrm{meeting}\mathrm{AB}\mathrm{at}\mathrm{L}\mathrm{and}\mathrm{M},\mathrm{respectively}. \\ \mathrm{Clearly},\mathrm{AMCD}\mathrm{is}\mathrm{a}\mathrm{parallelogram}. \\ \mathrm{Now}, \\ \mathrm{MC}=\mathrm{AD}=13\mathrm{cm} \\ \mathrm{AM}=\mathrm{DC}=11\mathrm{cm} \\ \Rightarrow \mathrm{MB}=\left(\mathrm{AB}-\mathrm{AM}\right) \end{gathered}$$
$$\begin{gathered} =\left(25-11\right)\mathrm{cm} \\ =14\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Thus},\mathrm{in}∆\mathrm{CMB},\mathrm{we}\mathrm{have}: \\ \mathrm{CM}=13\mathrm{cm} \\ \mathrm{MB}=14\mathrm{cm} \\ \mathrm{BC}=15\mathrm{cm} \\ \therefore \mathrm{s}=\frac{1}{2}\left(13+14+15\right)\mathrm{cm} \\ =\frac{1}{2}42\mathrm{cm} \\ =21\mathrm{cm} \\ \left(\mathrm{s}-\mathrm{a}\right)=\left(21-13\right)\mathrm{cm} \\ =8\mathrm{cm} \\ \left(\mathrm{s}-\mathrm{b}\right)=\left(21-14\right)\mathrm{cm} \\ =7\mathrm{cm} \\ \left(\mathrm{s}-\mathrm{c}\right)=\left(21-15\right)\mathrm{cm} \\ =6\mathrm{cm} \\ \therefore \mathrm{Area}\mathrm{of}∆\mathrm{CMB}=\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)} \end{gathered}$$
$$\begin{gathered} =\sqrt{21\times 8\times 7\times 6}{\mathrm{cm}}^2 \\ =84{\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \therefore \frac{1}{2}\times \mathrm{MB}\times \mathrm{CL}=84{\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times 14\times \mathrm{CL}=84{\mathrm{cm}}^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow \mathrm{CL}=\frac{84}{7} \\ \Rightarrow \mathrm{CL}=12\mathrm{cm} \end{gathered}$$
$\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{trapezium}=\left\{\frac{1}{2}\times \left(\mathrm{AB}+\mathrm{DC}\right)\times \mathrm{CL}\right\}$
$$\begin{gathered} =\left\{\frac{1}{2}\times \left(25+11\right)\times 12\right\}{\mathrm{cm}}^2 \\ =\left(\frac{1}{2}\times 36\times 12\right){\mathrm{cm}}^2 \\ =\left(18\times 12\right){\mathrm{cm}}^2 \\ =216{\mathrm{cm}}^2 \end{gathered}$$

$\mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{trapezium}\mathrm{is}216{\mathrm{cm}}^2.$