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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.

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Solution

## $\mathrm{Let}\mathrm{ABCD}\mathrm{be}\mathrm{the}\mathrm{trapezium}\mathrm{in}\mathrm{which}\mathrm{AB}\parallel \mathrm{DC},\mathrm{AB}=25\mathrm{cm},\mathrm{CD}=11\mathrm{cm},\mathrm{AD}=13\mathrm{cm}\mathrm{and}\mathrm{BC}=15\mathrm{cm}.\phantom{\rule{0ex}{0ex}}\mathrm{Draw}\mathrm{CL}\perp \mathrm{AB}\mathrm{and}\mathrm{CM}\parallel \mathrm{DA}\mathrm{meeting}\mathrm{AB}\mathrm{at}\mathrm{L}\mathrm{and}\mathrm{M},\mathrm{respectively}.\phantom{\rule{0ex}{0ex}}\mathrm{Clearly},\mathrm{AMCD}\mathrm{is}\mathrm{a}\mathrm{parallelogram}.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{MC}=\mathrm{AD}=13\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{AM}=\mathrm{DC}=11\mathrm{cm}\phantom{\rule{0ex}{0ex}}⇒\mathrm{MB}=\left(\mathrm{AB}-\mathrm{AM}\right)$ $=\left(25-11\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}=14\mathrm{cm}$ $\mathrm{Thus},\mathrm{in}∆\mathrm{CMB},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\mathrm{CM}=13\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{MB}=14\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{BC}=15\mathrm{cm}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{s}=\frac{1}{2}\left(13+14+15\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}42\mathrm{cm}\phantom{\rule{0ex}{0ex}}=21\mathrm{cm}\phantom{\rule{0ex}{0ex}}\left(\mathrm{s}-\mathrm{a}\right)=\left(21-13\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}=8\mathrm{cm}\phantom{\rule{0ex}{0ex}}\left(\mathrm{s}-\mathrm{b}\right)=\left(21-14\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}=7\mathrm{cm}\phantom{\rule{0ex}{0ex}}\left(\mathrm{s}-\mathrm{c}\right)=\left(21-15\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}=6\mathrm{cm}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Area}\mathrm{of}∆\mathrm{CMB}=\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ $=\sqrt{21×8×7×6}{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=84{\mathrm{cm}}^{2}$ $\therefore \frac{1}{2}×\mathrm{MB}×\mathrm{CL}=84{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}×14×\mathrm{CL}=84{\mathrm{cm}}^{2}$ $⇒\mathrm{CL}=\frac{84}{7}\phantom{\rule{0ex}{0ex}}⇒\mathrm{CL}=12\mathrm{cm}$ $\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{trapezium}=\left\{\frac{1}{2}×\left(\mathrm{AB}+\mathrm{DC}\right)×\mathrm{CL}\right\}$ $=\left\{\frac{1}{2}×\left(25+11\right)×12\right\}{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}×36×12\right){\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=\left(18×12\right){\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=216{\mathrm{cm}}^{2}$ $\mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{trapezium}\mathrm{is}216{\mathrm{cm}}^{2}.$

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