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Question-195750

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Question Number 195750 by sonukgindia last updated on 09/Aug/23
Commented by mr W last updated on 09/Aug/23
?=3×(8+8)−8−15=25
$$?=\mathrm{3}×\left(\mathrm{8}+\mathrm{8}\right)−\mathrm{8}−\mathrm{15}=\mathrm{25} \\ $$
Commented by SajaRashki last updated on 10/Aug/23
please explain your solution thank you
$${please}\:{explain}\:{your}\:{solution} \\ $$$${thank}\:{you} \\ $$
Answered by mr W last updated on 10/Aug/23
Commented by mr W last updated on 10/Aug/23
d=((a+b)/2) b=((c+d)/2) ⇒c=2b−d=2b−((a+b)/2)=((3b−a)/2) BE=(√3)b AD=(√3)d (((√3)ab)/2)=2×8 ⇒ab=((32)/( (√3))) (((√3)cd)/2)=2×15 ⇒cd=((60)/( (√3))) (((3b−a)/2))(((a+b)/2))=((60)/( (√3))) (3(√3)b^2 −(√3)ab)((√3)ab+(√3)b^2 )=240(√3)b^2 (3(√3)b^2 −32)(32+(√3)b^2 )=240(√3)b^2 let t=(√3)b^2 (3t−32)(t+32)=240t 3t^2 −176t−1024=0 ⇒t=64=(√3)b^2 ⇒b=(8/( (3)^(1/4) )) ⇒a=((32)/( (√3)b))=(4/( (3)^(1/4) )) ⇒(b/a)=2 Δ_(BCE) =2Δ_(ABE) ⇒Δ_(ACE) =3Δ_(ABE) =3×(8+8)=48 A_(red) =48−8−15=25 ✓
$${d}=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${b}=\frac{{c}+{d}}{\mathrm{2}}\:\Rightarrow{c}=\mathrm{2}{b}−{d}=\mathrm{2}{b}−\frac{{a}+{b}}{\mathrm{2}}=\frac{\mathrm{3}{b}−{a}}{\mathrm{2}} \\ $$$${BE}=\sqrt{\mathrm{3}}{b} \\ $$$${AD}=\sqrt{\mathrm{3}}{d} \\ $$$$\frac{\sqrt{\mathrm{3}}{ab}}{\mathrm{2}}=\mathrm{2}×\mathrm{8} \\ $$$$\Rightarrow{ab}=\frac{\mathrm{32}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{\sqrt{\mathrm{3}}{cd}}{\mathrm{2}}=\mathrm{2}×\mathrm{15} \\ $$$$\Rightarrow{cd}=\frac{\mathrm{60}}{\:\sqrt{\mathrm{3}}} \\ $$$$\left(\frac{\mathrm{3}{b}−{a}}{\mathrm{2}}\right)\left(\frac{{a}+{b}}{\mathrm{2}}\right)=\frac{\mathrm{60}}{\:\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{3}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} −\sqrt{\mathrm{3}}{ab}\right)\left(\sqrt{\mathrm{3}}{ab}+\sqrt{\mathrm{3}}{b}^{\mathrm{2}} \right)=\mathrm{240}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} −\mathrm{32}\right)\left(\mathrm{32}+\sqrt{\mathrm{3}}{b}^{\mathrm{2}} \right)=\mathrm{240}\sqrt{\mathrm{3}}{b}^{\mathrm{2}} \\ $$$${let}\:{t}=\sqrt{\mathrm{3}}{b}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}{t}−\mathrm{32}\right)\left({t}+\mathrm{32}\right)=\mathrm{240}{t} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{176}{t}−\mathrm{1024}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{64}=\sqrt{\mathrm{3}}{b}^{\mathrm{2}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{8}}{\:\sqrt[{\mathrm{4}}]{\mathrm{3}}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{32}}{\:\sqrt{\mathrm{3}}{b}}=\frac{\mathrm{4}}{\:\sqrt[{\mathrm{4}}]{\mathrm{3}}} \\ $$$$\Rightarrow\frac{{b}}{{a}}=\mathrm{2} \\ $$$$\Delta_{{BCE}} =\mathrm{2}\Delta_{{ABE}} \\ $$$$\Rightarrow\Delta_{{ACE}} =\mathrm{3}\Delta_{{ABE}} =\mathrm{3}×\left(\mathrm{8}+\mathrm{8}\right)=\mathrm{48} \\ $$$${A}_{{red}} =\mathrm{48}−\mathrm{8}−\mathrm{15}=\mathrm{25}\:\checkmark \\ $$

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