Let-f-be-a-real-valued-function-defined-on-the-inte-rval-1-1-If-the-area-of-the-equilateral-triangle-with-0-0-and-x-f-x-as-two-vertices-is-3-4-then-f-x-is-equal-to-A-1-x-2-

Question Number 119807 by Ar Brandon last updated on 27/Oct/20

$$\mathrm{Let}\:{f}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}-\mathrm{valued}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{on}\:\mathrm{the}\:\mathrm{inte}- \\ $$$$\mathrm{rval}\:\left[−\mathrm{1},\:\mathrm{1}\right].\:\mathrm{If}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\left(\mathrm{0},\:\mathrm{0}\right)\:\mathrm{and}\:\left(\mathrm{x},\:{f}\left(\mathrm{x}\right)\right)\:\mathrm{as}\:\mathrm{two}\:\mathrm{vertices}\:\mathrm{is}\:\sqrt{\mathrm{3}}/\mathrm{4},\:\mathrm{then}\:{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{A}\right)\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{C}\right)\:−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$

Answered by TANMAY PANACEA last updated on 28/Oct/20

side of equilatera [triangle=a=(√(x^2 +f^2 (x))) ((√3)/4)×a^2 =((√3)/4) a^2 =1 x^2 +f^2 (x)=1 f(x)=(√(1−x^2 ))

$${side}\:{of}\:{equilatera}\:\left[{triangle}={a}=\sqrt{{x}^{\mathrm{2}} +{f}^{\mathrm{2}} \left({x}\right)}\:\right. \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×{a}^{\mathrm{2}} \:=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${a}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{f}^{\mathrm{2}} \left({x}\right)=\mathrm{1} \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\: \\ $$

Commented by Ar Brandon last updated on 28/Oct/20

Thank you Sir

Commented by Ar Brandon last updated on 28/Oct/20

Thanks Sir

Commented by Ar Brandon last updated on 03/Nov/20

Sir how come ((√3)/4)×a^2 ? Is it by using (1/2)a×a sin60° ? 60° being the angle between two adjacent sides.

$$\mathrm{Sir}\:\mathrm{how}\:\mathrm{come}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{a}^{\mathrm{2}} \:?\:\mathrm{Is}\:\mathrm{it}\:\mathrm{by}\:\mathrm{using} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}×\mathrm{a}\:\mathrm{sin60}°\:?\:\mathrm{60}°\:\mathrm{being}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{between}\:\mathrm{two}\:\mathrm{adjacent}\:\mathrm{sides}. \\ $$

Commented by TANMAY PANACEA last updated on 28/Oct/20

area=(√(s(s−a)(s−b)(s−c))) here a=b=c s=((a+b+c)/2)=((a+a+a)/2)=((3a)/2) (s−a)=((3a)/2)−a=(a/2) area=(√(((3a)/2)×(a/2)×(a/2)×(a/2))) =((√3)/4)a^2

$${area}=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\: \\ $$$${here}\:{a}={b}={c} \\ $$$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}}=\frac{{a}+{a}+{a}}{\mathrm{2}}=\frac{\mathrm{3}{a}}{\mathrm{2}} \\ $$$$\left({s}−{a}\right)=\frac{\mathrm{3}{a}}{\mathrm{2}}−{a}=\frac{{a}}{\mathrm{2}} \\ $$$${area}=\sqrt{\frac{\mathrm{3}{a}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \\ $$

Commented by TANMAY PANACEA last updated on 28/Oct/20

or mdthod area=(1/2)×a×h sin60^o =(h/a)→h=((√3)/2)a area=((√3)/2)×a×(1/2)×a=((√3)/4)a^2

$${or}\:{mdthod}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}×{a}×{h} \\ $$$${sin}\mathrm{60}^{{o}} \:=\frac{{h}}{{a}}\rightarrow{h}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{a} \\ $$$${area}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×{a}×\frac{\mathrm{1}}{\mathrm{2}}×{a}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \\ $$

Commented by TANMAY PANACEA last updated on 28/Oct/20

$${yes}\:{sir} \\ $$

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