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Question Number 43131 by ajfour last updated on 07/Sep/18

Commented by ajfour last updated on 07/Sep/18

Q.43128  (A possible solution)

$${Q}.\mathrm{43128}\:\:\left({A}\:{possible}\:{solution}\right) \\ $$

Answered by ajfour last updated on 07/Sep/18

Area=(y/2)[(√(64−(((15)/(2y))+(y/2))^2 )) +(√(121−(((85)/(2y))+(y/2))^2 )) ]  where y∈ [5,15] .

$${Area}=\frac{{y}}{\mathrm{2}}\left[\sqrt{\mathrm{64}−\left(\frac{\mathrm{15}}{\mathrm{2}{y}}+\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\sqrt{\mathrm{121}−\left(\frac{\mathrm{85}}{\mathrm{2}{y}}+\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\right] \\ $$$${where}\:{y}\in\:\left[\mathrm{5},\mathrm{15}\right]\:. \\ $$

Answered by ajfour last updated on 07/Sep/18

let presently OP = y = b  point A is intersection of  x^2 +y^2 =64   and  x^2 +(y−b)^2 =49  ⇒  b(2y_A −b)=15   y_A =((15)/(2b))+(b/2)    ⇒  x_A =(√(64−(((15)/(2b))+(b/2))^2 ))   ..(i)  point B is the intersection of  x^2 +y^2 =121   and  x^2 +(y−b)^2 =36  ⇒  b(2y_B −b)=85    y_B =((85)/(2b))+(b/2)  ⇒   x_B = (√(121−(((85)/(2b))+(b/2))^2 ))   Area = (b/2)(x_A +x_B )  Area = (b/2)[(√(64−(((15)/(2b))+(b/2))^2 ))                                 +(√(121−(((85)/(2b))+(b/2))^2 )) ]  Replacing b by y  Area = (y/2)[(√(64−(((15)/(2y))+(y/2))^2 ))                             +(√(121−(((85)/(2y))+(y/2))^2 )) ].

$${let}\:{presently}\:{OP}\:=\:{y}\:=\:\boldsymbol{{b}} \\ $$$${point}\:{A}\:{is}\:{intersection}\:{of} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{64}\:\:\:{and} \\ $$$${x}^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} =\mathrm{49} \\ $$$$\Rightarrow\:\:{b}\left(\mathrm{2}{y}_{{A}} −{b}\right)=\mathrm{15} \\ $$$$\:{y}_{{A}} =\frac{\mathrm{15}}{\mathrm{2}{b}}+\frac{{b}}{\mathrm{2}}\:\: \\ $$$$\Rightarrow\:\:{x}_{{A}} =\sqrt{\mathrm{64}−\left(\frac{\mathrm{15}}{\mathrm{2}{b}}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:\:..\left({i}\right) \\ $$$${point}\:{B}\:{is}\:{the}\:{intersection}\:{of} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{121}\:\:\:{and} \\ $$$${x}^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} =\mathrm{36} \\ $$$$\Rightarrow\:\:{b}\left(\mathrm{2}{y}_{{B}} −{b}\right)=\mathrm{85} \\ $$$$\:\:{y}_{{B}} =\frac{\mathrm{85}}{\mathrm{2}{b}}+\frac{{b}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{x}_{{B}} =\:\sqrt{\mathrm{121}−\left(\frac{\mathrm{85}}{\mathrm{2}{b}}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} }\: \\ $$$${Area}\:=\:\frac{\boldsymbol{{b}}}{\mathrm{2}}\left(\boldsymbol{{x}}_{\boldsymbol{{A}}} +\boldsymbol{{x}}_{\boldsymbol{{B}}} \right) \\ $$$${Area}\:=\:\frac{\boldsymbol{{b}}}{\mathrm{2}}\left[\sqrt{\mathrm{64}−\left(\frac{\mathrm{15}}{\mathrm{2}{b}}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt{\mathrm{121}−\left(\frac{\mathrm{85}}{\mathrm{2}{b}}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\right] \\ $$$${Replacing}\:\boldsymbol{{b}}\:{by}\:{y} \\ $$$${Area}\:=\:\frac{{y}}{\mathrm{2}}\left[\sqrt{\mathrm{64}−\left(\frac{\mathrm{15}}{\mathrm{2}{y}}+\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt{\mathrm{121}−\left(\frac{\mathrm{85}}{\mathrm{2}{y}}+\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\right]. \\ $$$$ \\ $$

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