Question-213022

Question Number 213022 by Spillover last updated on 28/Oct/24

Answered by TonyCWX08 last updated on 31/Oct/24

FC = a+b height of trapezium_(AFBC) = (√(b^2 −((b/2))^2 )) = (√((3b^2 )/4)) = ((b(√3))/2) Area_(AFBC) =(1/2)(a+a+b)(((b(√3))/2)) =((2ab(√3)+b^2 (√3))/4) =((ab(√3))/2)+((b/2))^2 (√3) =(√3)(((ab)/2)+((b^2 /2))) Little equilateral triangle formed when joining point A to D and F to C Side length = b Area = (b^2 /4)(√3) Big equilateral triangle formed in the middle of the Hexagon when joining the same points Side length = a+b−2b = a−b Area = (((a−b)^2 )/4)(√3) Area of Cyclic Hexagon = 3Area_(AFBC) − 3Area_(SmallΔ) + Area_(BigΔ) = 3((√3)(((ab)/2)+(b^2 /4))) − 3((b^2 /4)(√3)) + ((a^2 −2ab+b^2 )/4)(√3) =((3ab(√3))/2)+((3(√3)b^2 )/4)−((3b^2 (√3))/4)+(((√3)(a^2 −2ab+b^2 ))/4) =((6ab(√3)+3b^2 (√3)−3b^2 (√3)+a^2 (√3)−2ab(√3)+b^2 (√3))/4) =((a^2 (√3)+4ab(√3)+b^2 (√3))/4) =(√3)[(a^2 /4)+ab+(b^2 /4)] =(√3)[((a/2))^2 +((b/2))^2 +ab] Hence Proved.

$${FC}\:=\:{a}+{b} \\ $$$${height}\:{of}\:{trapezium}_{{AFBC}} \: \\ $$$$=\:\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\:\sqrt{\frac{\mathrm{3}{b}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$=\:\frac{{b}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${Area}_{{AFBC}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{a}+{b}\right)\left(\frac{{b}\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{2}{ab}\sqrt{\mathrm{3}}+{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\frac{{ab}\sqrt{\mathrm{3}}}{\mathrm{2}}+\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} \sqrt{\mathrm{3}} \\ $$$$=\sqrt{\mathrm{3}}\left(\frac{{ab}}{\mathrm{2}}+\left(\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\right)\right) \\ $$$$ \\ $$$${Little}\:{equilateral}\:{triangle}\:{formed}\:{when}\:{joining}\:{point}\:{A}\:{to}\:{D}\:{and}\:{F}\:{to}\:{C} \\ $$$${Side}\:{length}\:=\:{b} \\ $$$${Area}\:=\:\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${Big}\:{equilateral}\:{triangle}\:{formed}\:{in}\:{the}\:{middle}\:{of}\:{the}\:{Hexagon}\:{when}\:{joining}\:{the}\:{same}\:{points} \\ $$$${Side}\:{length}\: \\ $$$$=\:{a}+{b}−\mathrm{2}{b} \\ $$$$=\:{a}−{b} \\ $$$${Area}\:=\:\frac{\left({a}−{b}\right)^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${Area}\:{of}\:{Cyclic}\:{Hexagon} \\ $$$$=\:\mathrm{3}{Area}_{{AFBC}} \:−\:\mathrm{3}{Area}_{{Small}\Delta} \:+\:{Area}_{{Big}\Delta} \\ $$$$=\:\mathrm{3}\left(\sqrt{\mathrm{3}}\left(\frac{{ab}}{\mathrm{2}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\right)\right)\:−\:\mathrm{3}\left(\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}}\right)\:+\:\frac{{a}^{\mathrm{2}} −\mathrm{2}{ab}+{b}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}} \\ $$$$=\frac{\mathrm{3}{ab}\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{3}{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}}+\frac{\sqrt{\mathrm{3}}\left({a}^{\mathrm{2}} −\mathrm{2}{ab}+{b}^{\mathrm{2}} \right)}{\mathrm{4}} \\ $$$$=\frac{\mathrm{6}{ab}\sqrt{\mathrm{3}}+\mathrm{3}{b}^{\mathrm{2}} \sqrt{\mathrm{3}}−\mathrm{3}{b}^{\mathrm{2}} \sqrt{\mathrm{3}}+{a}^{\mathrm{2}} \sqrt{\mathrm{3}}−\mathrm{2}{ab}\sqrt{\mathrm{3}}+{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}+\mathrm{4}{ab}\sqrt{\mathrm{3}}+{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\sqrt{\mathrm{3}}\left[\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+{ab}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\right] \\ $$$$=\sqrt{\mathrm{3}}\left[\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{ab}\right] \\ $$$${Hence}\:{Proved}. \\ $$

Answered by TonyCWX08 last updated on 31/Oct/24

Area_(PQR) =(X^2 /4)(√3) (X^2 /4)(√3) = (√3)[((b/2))^2 +((a/2))^2 +ab] ((X/2))^2 =[((a/2))^2 +ab+((b/2))^2 ] (X/2) = (√(((a/2))^2 +ab+((b/2))^2 )) (X/2) = (√((a^2 /4)+((4ab)/4)+(b^2 /4))) (X/2) = (√((a^2 +4ab+b^2 )/4)) (X/2) = ((√(a^2 +4ab+b^2 ))/2) X = (√(a^2 +4ab+b^2 ))

$${Area}_{{PQR}} =\frac{{X}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}} \\ $$$$\frac{{X}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}}\:=\:\sqrt{\mathrm{3}}\left[\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{ab}\right] \\ $$$$\left(\frac{{X}}{\mathrm{2}}\right)^{\mathrm{2}} =\left[\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{ab}+\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} \right] \\ $$$$\frac{{X}}{\mathrm{2}}\:=\:\sqrt{\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{ab}+\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\frac{{X}}{\mathrm{2}}\:=\:\:\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{4}{ab}}{\mathrm{4}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\frac{{X}}{\mathrm{2}}\:=\:\sqrt{\frac{{a}^{\mathrm{2}} +\mathrm{4}{ab}+{b}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\frac{{X}}{\mathrm{2}}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}{ab}+{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${X}\:=\:\sqrt{{a}^{\mathrm{2}} +\mathrm{4}{ab}+{b}^{\mathrm{2}} } \\ $$

Answered by TonyCWX08 last updated on 31/Oct/24