Question Number 154475 by mr W last updated on 18/Sep/21
Commented by mr W last updated on 18/Sep/21
$${find}\:{the}\:{area}\:{of}\:{the}\:{big}\:{triangle}\:{whose} \\ $$$${sides}\:{have}\:{the}\:{distances}\:\boldsymbol{{d}}_{\mathrm{1}} ,\boldsymbol{{d}}_{\mathrm{2}} ,\boldsymbol{{d}}_{\mathrm{3}} \:{to}\: \\ $$$${the}\:{sides}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\:{of}\:{the}\:{small}\:{triangle}. \\ $$
Commented by talminator2856791 last updated on 18/Sep/21
$$\:\mathrm{are}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{parallel}?\: \\ $$
Commented by mr W last updated on 18/Sep/21
$${yes},\:{it}\:{is}\:{said}. \\ $$
Commented by Tawa11 last updated on 21/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 19/Sep/21
$${let}\:\Delta={area}\:{of}\:{triangle}\:{ABC} \\ $$$$\Delta=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$${with}\:{s}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$$$ \\ $$$${say}\:{the}\:{incenter}\:{of}\:{triangle}\:{ABC}\:{is}\:{E} \\ $$$${and}\:{the}\:{radius}\:{of}\:{incircle}\:{is}\:{r}. \\ $$$${r}=\frac{\Delta}{{s}}=\frac{\mathrm{2}\Delta}{{a}+{b}+{c}} \\ $$$$ \\ $$$${say}\:{the}\:{distances}\:{from}\:{E}\:{to}\:{the}\:{sides} \\ $$$${of}\:{the}\:{triangle}\:{A}'{B}'{C}'\:{are}\:{h}_{\mathrm{1}} ,\:{h}_{\mathrm{2}} ,\:{h}_{\mathrm{3}} \\ $$$${respectively},\:{we}\:{have} \\ $$$${h}_{\mathrm{1}} ={r}+{d}_{\mathrm{1}} \\ $$$${h}_{\mathrm{2}} ={r}+{d}_{\mathrm{2}} \\ $$$${h}_{\mathrm{3}} ={r}+{d}_{\mathrm{3}} \\ $$
Commented by mr W last updated on 19/Sep/21
Commented by mr W last updated on 20/Sep/21
![it is obvious that both triangles are similar, since their sides are parallel to each other. say the side lengthes of the big triangle are a′, b′, c′ with a′=ka b′=kb c′=kc where k is the magnification factor. say the area of the big triangle A′B′C′ is Δ′. then we have Δ′=k^2 Δ. on the other side we have Δ′=((a′×h_1 )/2)+((b′×h_2 )/2)+((c′×h_3 )/2) Δ′=((ka×(r+d_1 ))/2)+((kb×(r+d_2 ))/2)+((kc×(r+d_3 ))/2) Δ′=(k/2)[(a+b+c)r+ad_1 +bd_2 +cd_3 ] Δ′=(k/2)[(a+b+c)((2Δ)/((a+b+c)))+ad_1 +bd_2 +cd_3 ] Δ′=(k/2)(2Δ+ad_1 +bd_2 +cd_3 ) since Δ′=k^2 Δ, (k/2)(2Δ+ad_1 +bd_2 +cd_3 )=k^2 Δ ⇒k=1+((ad_1 +bd_2 +cd_3 )/(2Δ)) therefore the area of big triangle is Δ′=(1+((ad_1 +bd_2 +cd_3 )/(2Δ)))^2 Δ](Q154511.png)
$${it}\:{is}\:{obvious}\:{that}\:{both}\:{triangles}\:{are} \\ $$$${similar},\:{since}\:{their}\:{sides}\:{are}\:{parallel} \\ $$$${to}\:{each}\:{other}. \\ $$$${say}\:{the}\:{side}\:{lengthes}\:{of}\:{the}\:{big} \\ $$$${triangle}\:{are}\:{a}',\:{b}',\:{c}'\:{with} \\ $$$${a}'={ka} \\ $$$${b}'={kb} \\ $$$${c}'={kc}\: \\ $$$${where}\:{k}\:{is}\:{the}\:{magnification}\:{factor}. \\ $$$$ \\ $$$${say}\:{the}\:{area}\:{of}\:{the}\:{big}\:{triangle}\:{A}'{B}'{C}' \\ $$$${is}\:\Delta'.\:{then}\:{we}\:{have}\:\Delta'={k}^{\mathrm{2}} \Delta. \\ $$$$ \\ $$$${on}\:{the}\:{other}\:{side}\:{we}\:{have} \\ $$$$\Delta'=\frac{{a}'×{h}_{\mathrm{1}} }{\mathrm{2}}+\frac{{b}'×{h}_{\mathrm{2}} }{\mathrm{2}}+\frac{{c}'×{h}_{\mathrm{3}} }{\mathrm{2}} \\ $$$$\Delta'=\frac{{ka}×\left({r}+{d}_{\mathrm{1}} \right)}{\mathrm{2}}+\frac{{kb}×\left({r}+{d}_{\mathrm{2}} \right)}{\mathrm{2}}+\frac{{kc}×\left({r}+{d}_{\mathrm{3}} \right)}{\mathrm{2}} \\ $$$$\Delta'=\frac{{k}}{\mathrm{2}}\left[\left({a}+{b}+{c}\right){r}+{ad}_{\mathrm{1}} +{bd}_{\mathrm{2}} +{cd}_{\mathrm{3}} \right] \\ $$$$\Delta'=\frac{{k}}{\mathrm{2}}\left[\left({a}+{b}+{c}\right)\frac{\mathrm{2}\Delta}{\left({a}+{b}+{c}\right)}+{ad}_{\mathrm{1}} +{bd}_{\mathrm{2}} +{cd}_{\mathrm{3}} \right] \\ $$$$\Delta'=\frac{{k}}{\mathrm{2}}\left(\mathrm{2}\Delta+{ad}_{\mathrm{1}} +{bd}_{\mathrm{2}} +{cd}_{\mathrm{3}} \right) \\ $$$${since}\:\Delta'={k}^{\mathrm{2}} \Delta, \\ $$$$\frac{{k}}{\mathrm{2}}\left(\mathrm{2}\Delta+{ad}_{\mathrm{1}} +{bd}_{\mathrm{2}} +{cd}_{\mathrm{3}} \right)={k}^{\mathrm{2}} \Delta \\ $$$$\Rightarrow{k}=\mathrm{1}+\frac{{ad}_{\mathrm{1}} +{bd}_{\mathrm{2}} +{cd}_{\mathrm{3}} }{\mathrm{2}\Delta} \\ $$$${therefore}\:{the}\:{area}\:{of}\:{big}\:{triangle}\:{is} \\ $$$$\Delta'=\left(\mathrm{1}+\frac{{ad}_{\mathrm{1}} +{bd}_{\mathrm{2}} +{cd}_{\mathrm{3}} }{\mathrm{2}\Delta}\right)^{\mathrm{2}} \Delta \\ $$
Commented by Rasheed.Sindhi last updated on 20/Sep/21
$$\mathbb{W}\mathrm{onderful}\:\mathbb{S}\mathrm{ir}! \\ $$
Commented by mr W last updated on 20/Sep/21
$${thanks}\:{for}\:{reviewing}\:{sir}! \\ $$
Answered by Rasheed.Sindhi last updated on 19/Sep/21
Commented by Rasheed.Sindhi last updated on 19/Sep/21
$$\bigtriangleup\mathrm{ABC}\sim\bigtriangleup\mathrm{A}'\mathrm{B}'\mathrm{C}' \\ $$$$\bigtriangleup\mathrm{ABC}\:{has}\:{been}\:{moved}\:{so}\:{that} \\ $$$$\mathrm{A}\:{coincide}\mathrm{s}\:\mathrm{A}' \\ $$$$\angle\mathrm{BAC}\:\:{coinsides}\:\angle\mathrm{B}'\mathrm{A}'\mathrm{C}' \\ $$$$\mathrm{d2}=\mathrm{d3}=\mathrm{0}\:{in}\:{this}\:{case}. \\ $$$${Continue} \\ $$
Commented by mr W last updated on 19/Sep/21
$$\frac{{ah}_{{a}} }{\mathrm{2}}=\Delta \\ $$$${h}_{{a}} =\frac{\mathrm{2}\Delta}{{a}} \\ $$$$\Delta'=\left(\frac{{h}_{{a}} +{d}_{\mathrm{1}} }{{h}_{{a}} }\right)^{\mathrm{2}} \Delta=\left(\mathrm{1}+\frac{{d}_{\mathrm{1}} }{\frac{\mathrm{2}\Delta}{{a}}}\right)^{\mathrm{2}} \Delta=\left(\mathrm{1}+\frac{{ad}_{\mathrm{1}} }{\mathrm{2}\Delta}\right)^{\mathrm{2}} \Delta \\ $$$$ \\ $$$${using}\:{my}\:{formula}\:{with}\:{d}_{\mathrm{2}} ={d}_{\mathrm{3}} =\mathrm{0} \\ $$$$\Delta'=\left(\mathrm{1}+\frac{{ad}_{\mathrm{1}} }{\mathrm{2}\Delta}\right)^{\mathrm{2}} \Delta\:\checkmark \\ $$
Commented by mr W last updated on 19/Sep/21
$${i}\:{have}\:{revised}\:{my}\:{solution}\:{above}. \\ $$$${please}\:{critical}\:{review}!\:{thanks}! \\ $$
Commented by Rasheed.Sindhi last updated on 19/Sep/21
$$\mathcal{T}{hanks}\:{to}\:{help}\:{for}\:{completing}\:{my} \\ $$$${approach}\:{also}. \\ $$