Question Number 223490 by mr W last updated on 26/Jul/25
Commented by Ghisom last updated on 27/Jul/25
$$\frac{\mathrm{1}}{\mathrm{2}}{a}\left({a}+{b}\right)−\frac{\mathrm{1}}{\mathrm{2}}{a}\left({a}−{b}\right)={ab}=\mathrm{77}{cm}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 27/Jul/25
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Answered by Raphael254 last updated on 27/Jul/25
$$ \\ $$$${Green}\:{area}\:{plus}\:{imaginary}\:{rectangle}\:{triangle}\:{area}\:=\:\frac{{x}\left({x}+{y}\right)}{\mathrm{2}} \\ $$$${Imaginary}\:{rectangle}\:{area}\:=\:\frac{{x}\left({x}−{y}\right)}{\mathrm{2}} \\ $$$${Green}\:{area}\:=\:\frac{{x}\left({x}+{y}\right)}{\mathrm{2}}\:−\:\frac{{x}\left({x}−{y}\right)}{\mathrm{2}}\:=\:\frac{{x}^{\mathrm{2}} \:+\:{xy}\:−{x}^{\mathrm{2}} \:+\:{xy}}{\mathrm{2}}\:=\:\frac{\mathrm{2}{xy}}{\mathrm{2}}\:=\:{xy}\:=\:\mathrm{77}\:{cm}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by Raphael254 last updated on 27/Jul/25
Commented by mr W last updated on 27/Jul/25
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Commented by fantastic last updated on 27/Jul/25
$${How}\:{did}\:{you}\:{make}\:{this} \\ $$$$??\:{please}\:{tell}\:{me} \\ $$$$ \\ $$
Answered by som(math1967) last updated on 27/Jul/25
Commented by som(math1967) last updated on 27/Jul/25
$${Area}\:{of}\:\bigtriangleup{ABD}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{77}{cm}^{\mathrm{2}} \\ $$$${Area}\:{of}\bigtriangleup\:{BCD}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{77}{cm}^{\mathrm{2}} \\ $$$${Green}\:{shaded}\:{area} \\ $$$$=\frac{\mathrm{77}}{\mathrm{2}}+\frac{\mathrm{77}}{\mathrm{2}}=\mathrm{77}{cm}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 27/Jul/25
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