Question Number 31409 by rahul 19 last updated on 08/Mar/18
$${The}\:{maximum}\:{area}\:{of}\:{the}\:{triangle} \\ $$$${whose}\:{sides}\:{a},{b}\:{and}\:{c}\:{satisfy}\: \\ $$$$\mathrm{0}\leqslant{a}\leqslant\mathrm{1}\:,\:\mathrm{1}\leqslant{b}\leqslant\mathrm{2}\:,\:\mathrm{2}\leqslant{c}\leqslant\mathrm{3}\:{is}\:: \\ $$$$\left.{A}\right)\:\mathrm{1} \\ $$$$\left.{B}\right)\:\mathrm{2} \\ $$$$\left.{C}\right)\:\mathrm{1}.\mathrm{5} \\ $$$$\left.{D}\right)\:\mathrm{0}.\mathrm{5}\:\:\:\:\:\:\:? \\ $$
Answered by MJS last updated on 08/Mar/18
$$\mathrm{I}\:\mathrm{think}\:\mathrm{because}\:\mathrm{of}\:{a}+{b}>{c},\:{a}+{c}>{b} \\ $$$$\mathrm{and}\:{b}+{c}>{a}\:\mathrm{and}\:{Area}=\frac{{s}×{h}_{{s}} }{\mathrm{2}}\:\mathrm{with} \\ $$$${s}={a}\vee{b}\vee{c}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{is}\:\mathrm{the}\:\mathrm{one} \\ $$$$\mathrm{with}\:\mathrm{max}\:{s}\:\mathrm{and}\:\mathrm{max}\:{h}_{{s}} \:\mathrm{and}\:\mathrm{it} \\ $$$$\mathrm{seems}\:\mathrm{to}\:\mathrm{me}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{rectangular} \\ $$$$\mathrm{one}\:\mathrm{with}\:{b}=\mathrm{2};\:{a}={h}_{{b}} =\mathrm{1};\:{c}=\sqrt{\mathrm{5}}\: \\ $$$$\mathrm{and}\:\frac{{b}×{h}_{{b}} }{\mathrm{2}}=\mathrm{1},\:\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{answer}\:\mathrm{1} \\ $$
Commented by rahul 19 last updated on 09/Mar/18
$${sir}\:{why}\:{u}\:{choose}\:{right}\:{angled}\:{triangle}. \\ $$$$\:{why}\:{not}\:{an}\:{acute}\:/\:{obtuse}\:{angled}\:{triangle} \\ $$
Commented by rahul 19 last updated on 09/Mar/18
$${your}\:{answer}\:{is}\:\:{correct}\:{but}\:{plz}\:{give}\:{reason}. \\ $$