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Question-73137




Question Number 73137 by behi83417@gmail.com last updated on 06/Nov/19
Commented by behi83417@gmail.com last updated on 06/Nov/19
AB^▲ C, is equilateral with: AB=a  BDFG,is rectangle.  find: S_(BH^▲ F) ,in terms of:  a .
$$\mathrm{A}\overset{\blacktriangle} {\mathrm{B}C},\:\mathrm{is}\:\mathrm{equilateral}\:\mathrm{with}:\:\mathrm{AB}=\mathrm{a} \\ $$$$\mathrm{BDFG},\mathrm{is}\:\mathrm{rectangle}. \\ $$$$\mathrm{find}:\:\mathrm{S}_{\mathrm{B}\overset{\blacktriangle} {\mathrm{H}F}} ,\mathrm{in}\:\mathrm{terms}\:\mathrm{of}:\:\:\mathrm{a}\:. \\ $$
Commented by mr W last updated on 06/Nov/19
Δ_(BHF)  is not constant.  question could be:  find max. Δ_(BHF)  in terms of a.
$$\Delta_{{BHF}} \:{is}\:{not}\:{constant}. \\ $$$${question}\:{could}\:{be}: \\ $$$${find}\:{max}.\:\Delta_{{BHF}} \:{in}\:{terms}\:{of}\:{a}. \\ $$
Commented by behi83417@gmail.com last updated on 07/Nov/19
hello dear mrW.  for what?please explain bit more if have  time.thank you.
$$\mathrm{hello}\:\mathrm{dear}\:\mathrm{mrW}. \\ $$$$\mathrm{for}\:\mathrm{what}?\mathrm{please}\:\mathrm{explain}\:\mathrm{bit}\:\mathrm{more}\:\mathrm{if}\:\mathrm{have} \\ $$$$\mathrm{time}.\mathrm{thank}\:\mathrm{you}. \\ $$
Commented by mr W last updated on 07/Nov/19
Commented by mr W last updated on 07/Nov/19
h=((√3)/2)x  A=(1/2)(a−x)((√3)/2)x=((√3)/4)(a−x)x ≠ constant  (dA/dx)=0 ⇒ x=(a/2) for A_(max)   A_(max) =((√3)/4)×(a/2)×(a/2)=(((√3)a^2 )/(16))
$${h}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{x}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({a}−{x}\right){x}\:\neq\:{constant} \\ $$$$\frac{{dA}}{{dx}}=\mathrm{0}\:\Rightarrow\:{x}=\frac{{a}}{\mathrm{2}}\:{for}\:{A}_{{max}} \\ $$$${A}_{{max}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{{a}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{16}} \\ $$
Commented by behi83417@gmail.com last updated on 07/Nov/19
thank you sir.  what happend if CD,be angular bisector  of ∡C?
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{what}\:\mathrm{happend}\:\mathrm{if}\:\mathrm{CD},\mathrm{be}\:\mathrm{angular}\:\mathrm{bisector} \\ $$$$\mathrm{of}\:\measuredangle\mathrm{C}? \\ $$
Commented by mr W last updated on 07/Nov/19
in that case, FG=DB=(a/( (√3))),  HB=CF=((FG)/(cos 30°))=((2a)/3)=a−x  ⇒x=(a/3)  A=((√3)/4)×(a/3)×((2a)/3)=(((√3)a^2 )/(18))
$${in}\:{that}\:{case},\:{FG}={DB}=\frac{{a}}{\:\sqrt{\mathrm{3}}}, \\ $$$${HB}={CF}=\frac{{FG}}{\mathrm{cos}\:\mathrm{30}°}=\frac{\mathrm{2}{a}}{\mathrm{3}}={a}−{x} \\ $$$$\Rightarrow{x}=\frac{{a}}{\mathrm{3}} \\ $$$${A}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{{a}}{\mathrm{3}}×\frac{\mathrm{2}{a}}{\mathrm{3}}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{18}} \\ $$
Commented by behi83417@gmail.com last updated on 07/Nov/19
dear master!thank you very much for  tring this.
$$\mathrm{dear}\:\mathrm{master}!\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{for} \\ $$$$\mathrm{tring}\:\mathrm{this}. \\ $$

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