Question Number 187205 by mr W last updated on 14/Feb/23
$${if}\:{the}\:{sides}\:{of}\:{a}\:{triangle}\:{are} \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} },\:\sqrt{{a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} },\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{respectively}, \\ $$$${what}\:{is}\:{its}\:{area}? \\ $$
Answered by ajfour last updated on 14/Feb/23
$${A}=\mathrm{4}{ab}−\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left({b}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left(\mathrm{2}{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left({b}\right)\left(\mathrm{2}{a}\right)\: \\ $$$${A}=\frac{{ab}}{\mathrm{2}}\left(\mathrm{8}−\mathrm{1}−\mathrm{2}−\mathrm{2}\right)=\frac{\mathrm{3}{ab}}{\mathrm{2}} \\ $$
Commented by anurup last updated on 14/Feb/23
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{by}\:\mathrm{using}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{formula}? \\ $$
Commented by ajfour last updated on 14/Feb/23
I wouldnt have dared..!
Commented by anurup last updated on 14/Feb/23
$$\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{too}\:\mathrm{cumbersome}\:\mathrm{I}\:\mathrm{guess}. \\ $$
Commented by mr W last updated on 15/Feb/23
$${nice}\:{solution}! \\ $$
Answered by ajfour last updated on 14/Feb/23
Answered by Frix last updated on 14/Feb/23
$$\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula}\:\mathrm{for}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{sides} \\ $$$${p},\:{q},\:{r}\:\mathrm{is} \\ $$$$\frac{\sqrt{\mathrm{2}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} {r}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} \right)−\left({p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{r}^{\mathrm{4}} \right)}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{for}\:\sqrt{{p}},\:\sqrt{{q}},\:\sqrt{{r}}\:\mathrm{it}\:\mathrm{is} \\ $$$$\frac{\sqrt{\mathrm{2}\left({pq}+{pr}+{qr}\right)−\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)}}{\mathrm{4}} \\ $$$${p}={u}+{v}\wedge{q}={u}+\mathrm{4}{v}\wedge{r}=\mathrm{4}{u}+{v} \\ $$$$\frac{\mathrm{3}\sqrt{{uv}}}{\mathrm{2}}=\frac{\mathrm{3}{ab}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 15/Feb/23
$${nice}\:{solution}! \\ $$