Question Number 224255 by fantastic last updated on 30/Aug/25
Commented by fantastic last updated on 30/Aug/25
$$\bigtriangleup{ABC}=? \\ $$
Commented by fantastic last updated on 30/Aug/25
$${r}_{\mathrm{1}} =\mathrm{5}\:,{r}_{\mathrm{2}} =\mathrm{20} \\ $$
Answered by mr W last updated on 30/Aug/25
Commented by mr W last updated on 30/Aug/25
$${BC}=\sqrt{\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{r}_{\mathrm{1}} {r}_{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\beta=\frac{\mathrm{2}\sqrt{{r}_{\mathrm{1}} {r}_{\mathrm{2}} }}{{r}_{\mathrm{1}} +{r}_{\mathrm{2}} }=\mathrm{sin}\:\alpha \\ $$$$\Delta{ABC}=\frac{\mathrm{2}\sqrt{{r}_{\mathrm{1}} {r}_{\mathrm{2}} }\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)}{\mathrm{2}}−\frac{\left({r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} \right)\mathrm{sin}\:\beta}{\mathrm{2}} \\ $$$$\:\:\:=\sqrt{{r}_{\mathrm{1}} {r}_{\mathrm{2}} }\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)−\frac{\sqrt{{r}_{\mathrm{1}} {r}_{\mathrm{2}} }\left({r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} \right)}{{r}_{\mathrm{1}} +{r}_{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{2}\left({r}_{\mathrm{1}} {r}_{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{r}_{\mathrm{1}} +{r}_{\mathrm{2}} }=\frac{\mathrm{2}\left(\mathrm{5}×\mathrm{20}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{5}+\mathrm{20}}=\mathrm{80}\:\checkmark \\ $$
Commented by fantastic last updated on 31/Aug/25
$${sir}\:{i}\:{have}\:{found}\:{two}\:{ways}. \\ $$$${one}\:{is}\:{very}\:{complex}\:{and}\:{other} \\ $$$${is}\:{easiest}\:{method}\left({i}\:{guess}\:{so}\right) \\ $$$${Thy}\:{both}\:{DO}\:{NOT}\:{use}\:{trigonometry} \\ $$
Commented by mr W last updated on 31/Aug/25
$${or} \\ $$$${BC}=\mathrm{2}\sqrt{{r}_{\mathrm{1}} {r}_{\mathrm{2}} } \\ $$$${AH}={r}_{\mathrm{1}} +\frac{{r}_{\mathrm{1}} \left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)}{{r}_{\mathrm{1}} +{r}_{\mathrm{2}} }=\frac{\mathrm{2}{r}_{\mathrm{1}} {r}_{\mathrm{2}} }{{r}_{\mathrm{1}} +{r}_{\mathrm{2}} } \\ $$$$\Delta{ABC}=\frac{{BC}×{AH}}{\mathrm{2}}=\frac{\mathrm{2}\left({r}_{\mathrm{1}} {r}_{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{r}_{\mathrm{1}} +{r}_{\mathrm{2}} }\:\checkmark \\ $$
Answered by fantastic last updated on 31/Aug/25
Commented by fantastic last updated on 31/Aug/25
$${let}\:{KL}={x}\: \\ $$$${so}\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{20}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}\sqrt{\mathrm{5}}×{x} \\ $$$${x}=\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\therefore{IF}=\mathrm{8}\sqrt{\mathrm{5}} \\ $$$$\bigtriangleup{IHF}=\mathrm{2}\bigtriangleup{IJF} \\ $$$$\bigtriangleup{IJF}=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}}\Rightarrow\mathrm{10}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\bigtriangleup{IJF}=\sqrt{\left(\mathrm{10}+\mathrm{4}\sqrt{\mathrm{5}}\right)\left(\mathrm{4}\sqrt{\mathrm{5}}\right)\left(\mathrm{4}\sqrt{\mathrm{5}}\right)\left(\mathrm{10}−\mathrm{4}\sqrt{\mathrm{5}}\right)} \\ $$$$=\mathrm{4}\sqrt{\mathrm{5}}×\sqrt{\mathrm{20}}=\mathrm{4}\sqrt{\mathrm{5}}×\mathrm{2}\sqrt{\mathrm{5}}=\mathrm{40} \\ $$$$\therefore\bigtriangleup{IHF}=\mathrm{2}×\mathrm{40}=\mathrm{80}{sq}.{units} \\ $$
Answered by fantastic last updated on 31/Aug/25
Commented by fantastic last updated on 31/Aug/25
$${another}\:{easy}\:{way} \\ $$