Question Number 7519 by Tawakalitu. last updated on 01/Sep/16
$${A}\:\left(\mathrm{2}\:×\:\mathrm{3}\right)\:{rectangle}\:{and}\:{a}\:\left(\mathrm{3}\:×\:\mathrm{4}\right)\:{rectangle}\:{are}\:{contain}\:{within}\: \\ $$$${a}\:{square}\:{without}\:{over}\:{laping}\:{at}\:{any}\:{inferior}\:{point}\:,\:{and}\:{the}\: \\ $$$${sides}\:{of}\:{the}\:{square}\:{are}\:{parallel}\:{to}\:{the}\:{sides}\:{of}\:{the}\:{two}\:{given} \\ $$$${rectangles}.\:{what}\:{is}\:{the}\:{smallest}\:{possible}\:{area}\:{of}\:{the}\:{square}. \\ $$
Commented by Rasheed Soomro last updated on 02/Sep/16
$$\mathrm{25} \\ $$
Commented by Yozzia last updated on 02/Sep/16
$${I}\:{think}\:\mathrm{5}^{\mathrm{2}} \:{also}. \\ $$
Commented by Rasheed Soomro last updated on 02/Sep/16
$${Sum}\:{of}\:{smaller}\:{sides}\:{of}\:{the}\:{rectangles} \\ $$$${should}\:{be}\:{measure}\:{of}\:{the}\:{side}\:{of}\:\:{required}\:{square} \\ $$$$\left({in}\:{case}\:{maximum}\:{of}\:{bigger}\:{sides}\:{is}\:{less}\:{or}\:{equal}\right. \\ $$$$\left.{to}\:{the}\:{mentioned}\:{sum}\right) \\ $$$${So}\:{in}\:{this}\:{case}\:\mathrm{2}+\mathrm{3}=\mathrm{5}\:{is}\:{the}\:{measure}\:{of}\:{the}\:{side} \\ $$$${of}\:{the}\:{square}. \\ $$$${Hence}\:{area}\:{of}\:{the}\:{square}\:{is}\:{of}\:{course}\:\mathrm{5}^{\mathrm{2}} =\mathrm{25}. \\ $$
Commented by Tawakalitu. last updated on 03/Sep/16
$${Thanks}\:{so}\:{much} \\ $$