Question-220262

Question Number 220262 by Tawa11 last updated on 10/May/25

Answered by mr W last updated on 10/May/25

Commented by mr W last updated on 10/May/25

y^2 =5^2 −x^2 3x^2 +5x^2 =8×(3×5+5^2 −x^2 ) ⇒x^2 =20 ⇒x=2(√5)

$${y}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{x}^{\mathrm{2}} =\mathrm{8}×\left(\mathrm{3}×\mathrm{5}+\mathrm{5}^{\mathrm{2}} −{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{20}\:\Rightarrow{x}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$

Commented by Tawa11 last updated on 10/May/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 10/May/25

Commented by mr W last updated on 10/May/25

h^2 =x^2 −(((5+3)/2))^2 =x^2 −4^2 y^2 =5^2 −x^2 h^2 =y^2 −(5−4)^2 =5^2 −x^2 −1^2 5^2 −x^2 −1^2 =x^2 −4^2 x^2 =20 ⇒x=2(√5)

$${h}^{\mathrm{2}} ={x}^{\mathrm{2}} −\left(\frac{\mathrm{5}+\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} ={y}^{\mathrm{2}} −\left(\mathrm{5}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \\ $$$$\mathrm{5}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\mathrm{20}\:\Rightarrow{x}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$

Commented by Tawa11 last updated on 10/May/25

God bless you sir. Please how do you get 1 sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{Please}\:\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\:\:\mathrm{1}\:\:\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 10/May/25