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