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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