Question Number 100644 by I want to learn more last updated on 27/Jun/20

Commented byghiniboss last updated on 27/Jun/20

looking for x?

Answered by mr W last updated on 28/Jun/20

let ∠ABP=α, ∠CBP=β  α+β=90°  15^2 =x^2 +9^2 −2x×9 cos α  ⇒cos α=((x^2 −144)/(18x))  12^2 =x^2 +9^2 −2x×9 cos β  ⇒cos β=((x^2 −63)/(18x))=sin α  (((x^2 −144)/(18x)))^2 +(((x^2 −63)/(18x)))^2 =1  2x^4 −738x^2 +24705=0  x^2 =((738±54(√(119)))/4)  x=(√((738+54(√(119)))/4))=18.214    there is an other solution, if it is  allowed that P can lie outside the  triangle:  x=(√((738−54(√(119)))/4))=6.101

Commented byI want to learn more last updated on 28/Jun/20

Thanks sir

Answered by 1549442205 last updated on 28/Jun/20

Commented by1549442205 last updated on 28/Jun/20

From the hypothesis we have:   { ((ABC^(�) =90°)),((BP=9 cm)),((PC=12 cm)),((PA=15 cm)) :}    we need to find x=AB=BC  Denoting by D,E be projections of P on  AB and BC respectively.Putting BD=m  AD=p,BE=n,CE=q,we have 9^2 −n^2 =PE^2 =12^2 −q^2  (1)  9^2 −m^2 =15^2 −p^2  (2).Since,m+p=n+q=x  ,from (1),(2) we get 63=x(q−n).From that  we get the system: { ((q+n=x)),((q−n=((63)/x))) :}    ⇒2n=x−((63)/x)  ⇒n=((x^2 −63)/(2x)).Similarly,we get m=((x^2 −144)/(2x))  On the other hands,m^2 +n^2 =9^2  ,so we have   (((x^2 −63)/(2x)))^2 +(((x^2 −144)/(2x)))^2 =81⇔2x^4 −414x^2 +24705=324x^2   ⇔2x^4 −738x^2 +24705=0.⇔x^2 ∈{331.7676;37.2323}  ⇔x∈{18.21 cm;6.10cm}

Commented byI want to learn more last updated on 28/Jun/20

Thanks sir

Answered by mr W last updated on 28/Jun/20

A(0,a)  B(0,0)  C(a,0)  P(k,h)  k^2 +h^2 =9^2    ...(i)  k^2 +(a−h)^2 =15^2    ...(ii)  (a−k)^2 +h^2 =12^2    ...(iii)    (ii)−(i):  a(a−2h)=144  ⇒h=(1/2)(a−((144)/a))   ...(iv)  (iii)−(i):  a(a−2k)=63  ⇒k=(1/2)(a−((63)/a))   ...(v)  put (iv) and (v) into (i):  (1/4)(a−((144)/a))^2 +(1/4)(a−((63)/a))^2 =81  (a−((144)/a))^2 +(a−((63)/a))^2 =324  2a^4 −738a^2 +24705=0  a^2 =((738±54(√(119)))/4)  ⇒x=a=(√((738±54(√(119)))/4))=18.214, 6.101

Commented byI want to learn more last updated on 28/Jun/20

Thanks sir