Question Number 224502 by fantastic last updated on 15/Sep/25
$${in}\:{a}\:{rhombus}\:{the}\:{perimeter} \\ $$$${is}\:\mathrm{2}{p}\:{and}\:{the}\:{sum}\:{of}\:{its}\:{diagonals} \\ $$$${is}\:{m}. \\ $$$${the}\:{area}\:{of}\:{rhombus}\:{is}… \\ $$
Answered by som(math1967) last updated on 15/Sep/25
$${each}\:{side}\:{of}\:{the}\:{rhombus}=\frac{{p}}{\mathrm{2}} \\ $$$$\:{let}\:{diagonals}\:{are}\:{x},{y} \\ $$$$\:{x}+{y}={m} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{y}^{\mathrm{2}} }{\mathrm{4}}+\frac{{xy}}{\mathrm{2}}=\frac{{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\:\frac{{p}^{\mathrm{2}} }{\mathrm{4}}\:+\frac{{xy}}{\mathrm{2}}=\frac{{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$$$\Rightarrow\frac{{xy}}{\mathrm{2}}=\frac{{m}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:={area}\:{of}\: \\ $$$${the}\:{rhombus} \\ $$
Commented by fantastic last updated on 15/Sep/25
$${sir} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \neq\left(\frac{{p}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$
Commented by som(math1967) last updated on 15/Sep/25
$${yes}\:{i}\:{have}\:{a}\:{mistake} \\ $$
Commented by fantastic last updated on 15/Sep/25
$${now}\:{it}\:{is}\:{right}! \\ $$