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Question Number 186551 by ajfour last updated on 05/Feb/23

Commented by mr W last updated on 06/Feb/23

square with side length 1? A_(shade) =((t^2 (t+2)^2 )/(4(1+t)[(1+t)^2 +1]))

squarewithsidelength1?Ashade=t2(t+2)24(1+t)[(1+t)2+1]

Answered by ajfour last updated on 06/Feb/23

Left bottom corner be Origin. q=(p/(1+t)) and q=(t+1)(1−p) ⇒ p=(((t+1)^2 )/(1+(t+1)^2 )) q=((t+1)/(1+(t+1)^2 )) k=((t+1)/2)−(1/(2(t+1))) ; h=p−(1/2) Area=kh =(((t+1)/2)−(1/(2(t+1))))((((t+1)^2 )/(1+(t+1)^2 ))−(1/2)) =(({(t+1)^2 −1})/(2(t+1)))×(({(t+1)^2 −1})/(2(t^2 +2t+2))) A=(({(t+1)^2 −1}^2 )/(4(t+1)(t^2 +2t+2))) A=((t^2 (t+2)^2 )/(4(t+1)(t^2 +2t+2)))

LeftbottomcornerbeOrigin.q=p1+tandq=(t+1)(1−p)⇒p=(t+1)21+(t+1)2q=t+11+(t+1)2k=t+12−12(t+1);h=p−12Area=kh=(t+12−12(t+1))((t+1)21+(t+1)2−12)={(t+1)2−1}2(t+1)×{(t+1)2−1}2(t2+2t+2)A={(t+1)2−1}24(t+1)(t2+2t+2)A=t2(t+2)24(t+1)(t2+2t+2)

Answered by mr W last updated on 06/Feb/23

Commented by mr W last updated on 06/Feb/23

h_2 =((1+t)/2) h_1 =(1/(2(1+t))) ((1−b)/h_3 )=((1+t)/1) ⇒(1/h_3 )=((1+t)/(1−b)) (b/h_3 )=(1/(1+t)) ⇒(1/h_3 )=(1/(b(1+t))) ((1+t)/(1−b))=(1/(b(1+t))) ⇒b=(1/((1+t)^2 +1)) A_(shade) =(h_2 −h_1 )((1/2)−b) =[((1+t)/2)−(1/(2(1+t)))][(1/2)−(1/((1+t)^2 +1))] =((t^2 (t+2)^2 )/(4(1+t)(t^2 +2t+2)))

h2=1+t2h1=12(1+t)1−bh3=1+t1⇒1h3=1+t1−bbh3=11+t⇒1h3=1b(1+t)1+t1−b=1b(1+t)⇒b=1(1+t)2+1Ashade=(h2−h1)(12−b)=[1+t2−12(1+t)][12−1(1+t)2+1]=t2(t+2)24(1+t)(t2+2t+2)

Commented by ajfour last updated on 06/Feb/23

Thanks you Sir, I corrected!

ThanksyouSir,Icorrected!

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