Question Number 224927 by behi834171 last updated on 12/Oct/25
Commented by behi834171 last updated on 12/Oct/25
$$\boldsymbol{{s}}_{\mathrm{1}} =\mathrm{1};\boldsymbol{{s}}_{\mathrm{2}} =\mathrm{2};\boldsymbol{{s}}_{\mathrm{3}} =\mathrm{3}\left(\boldsymbol{{unit}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{S}}_{\bigtriangleup} =?\:\left(\boldsymbol{{unit}}\right)^{\mathrm{2}} \\ $$
Commented by mr W last updated on 14/Oct/25
$${see}\:{Q}\mathrm{83268} \\ $$$${S}=\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AC}} \\ $$$$\:\:\:=\sqrt{\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\mathrm{3}}=\mathrm{4}.\mathrm{899} \\ $$
Answered by som(math1967) last updated on 13/Oct/25
![let length of rectangle=xunit breadth =y unit (1/2)(x−(6/y))(y−(2/x))=S_2 ⇒ xy−2−6+((12)/(xy))=2×2 ⇒xy+((12)/(xy))=12 ⇒a^2 −12a+12=0 [xy=a=area of rect] ⇒a=((12±(√(144−48)))/2) ∴a=10.89 or1.1 but a>S_2 ∴ a=10.89squnit ∴ S_4 =10.89−(1+2+3)=4.89squnit](https://www.tinkutara.com/question/Q224934.png)
$${let}\:{length}\:{of}\:{rectangle}={xunit} \\ $$$$\:\:{breadth}\:={y}\:{unit} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{6}}{{y}}\right)\left({y}−\frac{\mathrm{2}}{{x}}\right)={S}_{\mathrm{2}} \\ $$$$\Rightarrow\:{xy}−\mathrm{2}−\mathrm{6}+\frac{\mathrm{12}}{{xy}}=\mathrm{2}×\mathrm{2} \\ $$$$\Rightarrow{xy}+\frac{\mathrm{12}}{{xy}}=\mathrm{12} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{12}=\mathrm{0}\:\left[{xy}={a}={area}\:{of}\:{rect}\right] \\ $$$$\Rightarrow{a}=\frac{\mathrm{12}\pm\sqrt{\mathrm{144}−\mathrm{48}}}{\mathrm{2}} \\ $$$$\:\therefore{a}=\mathrm{10}.\mathrm{89}\:{or}\mathrm{1}.\mathrm{1} \\ $$$$\:{but}\:{a}>{S}_{\mathrm{2}} \:\therefore\:{a}=\mathrm{10}.\mathrm{89}{squnit} \\ $$$$\therefore\:{S}_{\mathrm{4}} =\mathrm{10}.\mathrm{89}−\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)=\mathrm{4}.\mathrm{89}{squnit} \\ $$