Question Number 17948 by b.e.h.i.8.3.417@gmail.com last updated on 12/Jul/17
Commented by mrW1 last updated on 12/Jul/17
![area of ΔABC=S=(1/2)ah_a ⇒(1/h_a )=(1/(2S)) a=ka similarly ⇒(1/h_b )=kb ⇒(1/h_c )=kc ⇒the triangle with sides (1/h_a ),(1/h_b ),(1/h_c ) is similary to the triangle with sides a,b,c. the ratio of their side lengthes is k=(1/(2S)). area of new triangle is S′ S′=k^2 S=(1/(4S^2 )) S=(1/(4S)) ⇒S∙S′=(1/4) we use k=(1/(2S)) only as dimensionless factor for the side lengthes of the new triangle: a′=k a b′=k b c′ =k c a′,b′,c′ have the same unit as a,b,c, e.g. [cm]. also S′ has the same unit as S, e.g. [cm^2 ] S∙S′=(1/4) means only that the quantity is (1/4), the unit is [cm^4 ]. for example: triangle ΔABC with sides a=3 cm b=4 cm c=5 cm its area S=(1/2)×3×4=6 cm^2 now we construct a new triangle with side lengthes which are k−times of the side lengthes from ΔABC. k=(1/(2S))=(1/(2×6))=(1/(12)) (only as number w/o unit) a′=ka=(3/(12)) cm b′=kb=(4/(12)) cm c′=kc=(5/(12)) cm the area of this triangle is S′=(1/2)×(3/(12))×(4/(12))=(1/(24)) cm^2 S×S′=6×(1/(24))=(1/4) cm^4](https://www.tinkutara.com/question/Q17953.png)
$$\mathrm{area}\:\mathrm{of}\:\Delta\mathrm{ABC}=\mathrm{S}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ah}_{\mathrm{a}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{a}} }=\frac{\mathrm{1}}{\mathrm{2S}}\:\mathrm{a}=\mathrm{ka} \\ $$$$\mathrm{similarly} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{b}} }=\mathrm{kb}\: \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{c}} }=\mathrm{kc}\: \\ $$$$\Rightarrow\mathrm{the}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{sides}\:\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{a}} },\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{b}} },\frac{\mathrm{1}}{\mathrm{h}_{\mathrm{c}} } \\ $$$$\mathrm{is}\:\mathrm{similary}\:\mathrm{to}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{sides} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}.\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{their}\:\mathrm{side}\:\mathrm{lengthes} \\ $$$$\mathrm{is}\:\mathrm{k}=\frac{\mathrm{1}}{\mathrm{2S}}. \\ $$$$ \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{new}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{S}' \\ $$$$\mathrm{S}'=\mathrm{k}^{\mathrm{2}} \:\mathrm{S}=\frac{\mathrm{1}}{\mathrm{4S}^{\mathrm{2}} }\:\mathrm{S}=\frac{\mathrm{1}}{\mathrm{4S}} \\ $$$$\Rightarrow\mathrm{S}\centerdot\mathrm{S}'=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{use}\:\mathrm{k}=\frac{\mathrm{1}}{\mathrm{2S}}\:\mathrm{only}\:\mathrm{as}\:\mathrm{dimensionless}\:\mathrm{factor} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{side}\:\mathrm{lengthes}\:\mathrm{of}\:\mathrm{the}\:\mathrm{new}\:\mathrm{triangle}: \\ $$$$\mathrm{a}'=\mathrm{k}\:\mathrm{a} \\ $$$$\mathrm{b}'=\mathrm{k}\:\mathrm{b} \\ $$$$\mathrm{c}'\:=\mathrm{k}\:\mathrm{c} \\ $$$$\mathrm{a}',\mathrm{b}',\mathrm{c}'\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{unit}\:\mathrm{as}\:\mathrm{a},\mathrm{b},\mathrm{c},\:\mathrm{e}.\mathrm{g}.\:\left[\mathrm{cm}\right]. \\ $$$$\mathrm{also}\:\mathrm{S}'\:\mathrm{has}\:\mathrm{the}\:\mathrm{same}\:\mathrm{unit}\:\mathrm{as}\:\mathrm{S},\:\mathrm{e}.\mathrm{g}.\:\left[\mathrm{cm}^{\mathrm{2}} \right] \\ $$$$\mathrm{S}\centerdot\mathrm{S}'=\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{means}\:\mathrm{only}\:\mathrm{that}\:\mathrm{the}\:\mathrm{quantity}\:\mathrm{is} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}},\:\mathrm{the}\:\mathrm{unit}\:\mathrm{is}\:\left[\mathrm{cm}^{\mathrm{4}} \right]. \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{example}: \\ $$$$\mathrm{triangle}\:\Delta\mathrm{ABC}\:\mathrm{with}\:\mathrm{sides} \\ $$$$\mathrm{a}=\mathrm{3}\:\mathrm{cm} \\ $$$$\mathrm{b}=\mathrm{4}\:\mathrm{cm} \\ $$$$\mathrm{c}=\mathrm{5}\:\mathrm{cm} \\ $$$$\mathrm{its}\:\mathrm{area}\:\mathrm{S}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{3}×\mathrm{4}=\mathrm{6}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{construct}\:\mathrm{a}\:\mathrm{new}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\mathrm{side}\:\mathrm{lengthes}\:\mathrm{which}\:\mathrm{are}\:\mathrm{k}−\mathrm{times}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{side}\:\mathrm{lengthes}\:\mathrm{from}\:\Delta\mathrm{ABC}. \\ $$$$\mathrm{k}=\frac{\mathrm{1}}{\mathrm{2S}}=\frac{\mathrm{1}}{\mathrm{2}×\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{12}}\:\left(\mathrm{only}\:\mathrm{as}\:\mathrm{number}\:\mathrm{w}/\mathrm{o}\:\mathrm{unit}\right) \\ $$$$\mathrm{a}'=\mathrm{ka}=\frac{\mathrm{3}}{\mathrm{12}}\:\mathrm{cm} \\ $$$$\mathrm{b}'=\mathrm{kb}=\frac{\mathrm{4}}{\mathrm{12}}\:\mathrm{cm} \\ $$$$\mathrm{c}'=\mathrm{kc}=\frac{\mathrm{5}}{\mathrm{12}}\:\mathrm{cm} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{this}\:\mathrm{triangle}\:\mathrm{is} \\ $$$$\mathrm{S}'=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{12}}×\frac{\mathrm{4}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{24}}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$\mathrm{S}×\mathrm{S}'=\mathrm{6}×\frac{\mathrm{1}}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{cm}^{\mathrm{4}} \\ $$
Commented by b.e.h.i.8.3.417@gmail.com last updated on 13/Jul/17
$${thank}\:{you}\:{very}\:{much}\:{dear}\:{master}. \\ $$$${how}\:{can}\:{we}\:{draw}\:{this}\:{triangle}?{or}\:{show}\:{it} \\ $$$${in}\:{ABC}'\:{plan}? \\ $$
Commented by mrW1 last updated on 13/Jul/17
$$\mathrm{to}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{new}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{just}\:\mathrm{to} \\ $$$$\mathrm{reduce}\:\mathrm{the}\:\mathrm{original}\:\mathrm{triangle}\:\mathrm{in}\:\mathrm{ratio} \\ $$$$\mathrm{1}:\mathrm{2S}. \\ $$
Answered by wullyfalcon last updated on 12/Jul/17
$${the}\:{triangle}\:{is}\:{equal}\:{on}\:{all}\:{sides}\: \\ $$$${because}\:{the}\:{triangle}\:{is}\:{in}\:{space}\: \\ $$$${it}\:{can}\:{be}\:{any}\:{size}\:{a}={b}\:{b}={a}\:{a}={c}\:{c}={b}\:{ect} \\ $$$${draw}\:{a}\:{triangle}\:{with}\:\mathrm{3}\:{equal}\:{sides} \\ $$$${the}\:{value}\:{for}\:{this}\:{triangle}\:{is}\:\infty \\ $$