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Question Number 174591 by Mastermind last updated on 05/Aug/22
The drawing below shows two equilateral triangles with side length a. The triangle are horizontally shifted by (a/2). Find the intersection area A of the two triangles (grey area).
$$\mathrm{The}\:\mathrm{drawing}\:\mathrm{below}\:\mathrm{shows}\:\mathrm{two}\: \\ $$$$\mathrm{equilateral}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{side}\:\mathrm{length} \\ $$$$\boldsymbol{\mathrm{a}}.\:\mathrm{The}\:\mathrm{triangle}\:\mathrm{are}\:\mathrm{horizontally}\:\mathrm{shifted} \\ $$$$\mathrm{by}\:\frac{\mathrm{a}}{\mathrm{2}}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{area}\:\mathrm{A}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{triangles}\:\left(\mathrm{grey}\:\mathrm{area}\right). \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by behi834171 last updated on 05/Aug/22
aslo have an equilateral triangle with side lenght: (a/2) so,wanted area=((((a/2))^2 )/4)(√3)=((a^2 (√3))/(16)) .
$$\boldsymbol{{aslo}}\:\boldsymbol{{have}}\:\boldsymbol{{an}}\:\boldsymbol{{equilateral}}\:\boldsymbol{{triangle}} \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{side}}\:\boldsymbol{{lenght}}:\:\frac{\boldsymbol{{a}}}{\mathrm{2}} \\ $$$$\boldsymbol{{so}},\boldsymbol{{wanted}}\:\boldsymbol{{area}}=\frac{\left(\frac{\boldsymbol{{a}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}}=\frac{\boldsymbol{{a}}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{16}}\:\:. \\ $$
Commented by Mastermind last updated on 05/Aug/22
I did not understand what you′re trying to say here, explain it well
$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{understand}\:\mathrm{what}\:\mathrm{you}'\mathrm{re}\:\mathrm{trying} \\ $$$$\mathrm{to}\:\mathrm{say}\:\mathrm{here},\:\mathrm{explain}\:\mathrm{it}\:\mathrm{well} \\ $$
Commented by Mastermind last updated on 05/Aug/22
Commented by behi834171 last updated on 05/Aug/22
4A=((a^2 (√3))/4)⇒A=((a^2 (√3))/(16)) . 4 of such shaded areas =whole of a equilateral triangle with side: a that it′s area= ((a^2 (√3))/4) . i got what you meant sir . it is too easy to say more ....
$$\:\:\:\:\:\:\:\:\:\mathrm{4}\boldsymbol{{A}}=\frac{\boldsymbol{{a}}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}}\Rightarrow\boldsymbol{{A}}=\frac{\boldsymbol{{a}}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{16}}\:\:\:. \\ $$$$\mathrm{4}\:\boldsymbol{{of}}\:\boldsymbol{{such}}\:\boldsymbol{{shaded}}\:\boldsymbol{{areas}}\:=\boldsymbol{{whole}}\:\:\boldsymbol{{of}}\:\boldsymbol{{a}} \\ $$$$\boldsymbol{{equilateral}}\:\boldsymbol{{triangle}}\:\boldsymbol{{with}}\:\boldsymbol{{side}}:\:\boldsymbol{{a}} \\ $$$$\boldsymbol{{that}}\:\boldsymbol{{it}}'\boldsymbol{{s}}\:\boldsymbol{{area}}=\:\frac{\boldsymbol{{a}}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}}\:\:. \\ $$$$ \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{got}}\:\boldsymbol{{what}}\:\boldsymbol{{you}}\:\boldsymbol{{meant}}\:\boldsymbol{{sir}}\:. \\ $$$$\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{too}}\:\boldsymbol{{easy}}\:\boldsymbol{{to}}\:\boldsymbol{{say}}\:\boldsymbol{{more}}\:…. \\ $$
Commented by Mastermind last updated on 05/Aug/22
Tank you so much man
$$\mathrm{Tank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{man} \\ $$

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