Question Number 206783 by BaliramKumar last updated on 25/Apr/24
Commented by BaliramKumar last updated on 25/Apr/24
$$\mathrm{i}\:\mathrm{find}\:\mathrm{20} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Apr/24
$${let}\:{along}\:{the}\:{length}\:{are}\:{m}\:{squares} \\ $$$$\&\:{along}\:{the}\:{width}\:{are}\:{n}\:{squares}. \\ $$$${Side}\:{of}\:{a}\:{square}=\frac{\mathrm{4}.\mathrm{25}}{{m}}=\frac{\mathrm{3}.\mathrm{4}}{{n}} \\ $$$$\Rightarrow\frac{{m}}{{n}}=\frac{\mathrm{4}.\mathrm{25}}{\mathrm{3}.\mathrm{4}}=\frac{\mathrm{425}}{\mathrm{340}}=\frac{\mathrm{5}}{\mathrm{4}}\:\left(\because\:{m},{n}\in\mathbb{N}\right) \\ $$$${m}_{\mathrm{min}} =\mathrm{5},{n}_{\mathrm{min}} =\mathrm{4} \\ $$$${Number}\:{of}\:{squares}\:{along}\:{length} \\ $$$$=\mathrm{5} \\ $$$${Number}\:{of}\:{squares}\:{along}\:{width} \\ $$$$=\mathrm{4} \\ $$$$\mathcal{T}{otal}\:{number}\left({least}\right)\:{of}\:{squares}=\mathrm{20} \\ $$$${When}\:{the}\:{number}\:{of}\:{squares} \\ $$$${is}\:{least}\:{the}\:{area}\:{of}\:{square}\:{is}\:{largest}. \\ $$$$ \\ $$