Question-207018

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Question Number 207018 by cherokeesay last updated on 03/May/24

Answered by mr W last updated on 03/May/24

Commented by cherokeesay last updated on 03/May/24

$${R}=? \\ $$

Commented by mr W last updated on 03/May/24

4z^2 =2×10^2 +2×9^2 −(2R)^2 ⇒z^2 =((181)/2)−R^2 BF^2 =x^2 +R^2 CF^2 =y^2 +R^2 4z^2 +10(x^2 +R^2 )=14(R^2 +4×10) ⇒4z^2 +10x^2 =4R^2 +560 ⇒5x^2 =4R^2 +99 5z^2 +9(y^2 +R^2 )=14(R^2 +5×9) ⇒5z^2 +9y^2 =5R^2 +630 ⇒18y^2 =20R^2 +355 cos A=((14^2 +14^2 −(x+y)^2 )/(2×14×14))=((10^2 +9^2 −(2R)^2 )/(2×10×9)) ⇒45(x+y)^2 =392R^2 −98 ⇒45(x^2 +y^2 +2xy)=392R^2 −98 ⇒20xy=68R^2 −417 ⇒20^2 x^2 y^2 =(68R^2 −417)^2 ⇒20^2 ×((4R^2 +99)/5)×((20R^2 +355)/(18))=(68R^2 −417)^2 ⇒784R^4 −13192R^2 +3249=0 ⇒R=(√((6596+(√(6596^2 −784×3249)))/(784)))=((57)/(14))≈4.071

$$\mathrm{4}{z}^{\mathrm{2}} =\mathrm{2}×\mathrm{10}^{\mathrm{2}} +\mathrm{2}×\mathrm{9}^{\mathrm{2}} −\left(\mathrm{2}{R}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{z}^{\mathrm{2}} =\frac{\mathrm{181}}{\mathrm{2}}−{R}^{\mathrm{2}} \\ $$$${BF}^{\mathrm{2}} ={x}^{\mathrm{2}} +{R}^{\mathrm{2}} \\ $$$${CF}^{\mathrm{2}} ={y}^{\mathrm{2}} +{R}^{\mathrm{2}} \\ $$$$\mathrm{4}{z}^{\mathrm{2}} +\mathrm{10}\left({x}^{\mathrm{2}} +{R}^{\mathrm{2}} \right)=\mathrm{14}\left({R}^{\mathrm{2}} +\mathrm{4}×\mathrm{10}\right) \\ $$$$\Rightarrow\mathrm{4}{z}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} +\mathrm{560} \\ $$$$\Rightarrow\mathrm{5}{x}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} +\mathrm{99} \\ $$$$\mathrm{5}{z}^{\mathrm{2}} +\mathrm{9}\left({y}^{\mathrm{2}} +{R}^{\mathrm{2}} \right)=\mathrm{14}\left({R}^{\mathrm{2}} +\mathrm{5}×\mathrm{9}\right) \\ $$$$\Rightarrow\mathrm{5}{z}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} =\mathrm{5}{R}^{\mathrm{2}} +\mathrm{630} \\ $$$$\Rightarrow\mathrm{18}{y}^{\mathrm{2}} =\mathrm{20}{R}^{\mathrm{2}} +\mathrm{355} \\ $$$$\mathrm{cos}\:{A}=\frac{\mathrm{14}^{\mathrm{2}} +\mathrm{14}^{\mathrm{2}} −\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{14}×\mathrm{14}}=\frac{\mathrm{10}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\left(\mathrm{2}{R}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{10}×\mathrm{9}} \\ $$$$\Rightarrow\mathrm{45}\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{392}{R}^{\mathrm{2}} −\mathrm{98} \\ $$$$\Rightarrow\mathrm{45}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}\right)=\mathrm{392}{R}^{\mathrm{2}} −\mathrm{98} \\ $$$$\Rightarrow\mathrm{20}{xy}=\mathrm{68}{R}^{\mathrm{2}} −\mathrm{417} \\ $$$$\Rightarrow\mathrm{20}^{\mathrm{2}} {x}^{\mathrm{2}} {y}^{\mathrm{2}} =\left(\mathrm{68}{R}^{\mathrm{2}} −\mathrm{417}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{20}^{\mathrm{2}} ×\frac{\mathrm{4}{R}^{\mathrm{2}} +\mathrm{99}}{\mathrm{5}}×\frac{\mathrm{20}{R}^{\mathrm{2}} +\mathrm{355}}{\mathrm{18}}=\left(\mathrm{68}{R}^{\mathrm{2}} −\mathrm{417}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{784}{R}^{\mathrm{4}} −\mathrm{13192}{R}^{\mathrm{2}} +\mathrm{3249}=\mathrm{0} \\ $$$$\Rightarrow{R}=\sqrt{\frac{\mathrm{6596}+\sqrt{\mathrm{6596}^{\mathrm{2}} −\mathrm{784}×\mathrm{3249}}}{\mathrm{784}}}=\frac{\mathrm{57}}{\mathrm{14}}\approx\mathrm{4}.\mathrm{071} \\ $$

Answered by mr W last updated on 04/May/24

Commented by mr W last updated on 03/May/24

((4 cos θ+5 cos θ)/2)=R ⇒cos θ=((2R)/9) (2R)^2 =10^2 +9^2 −2×10×9 cos 2θ 4R^2 =181−180(2×((4R^2 )/(81))−1) R^2 =((3249)/(196)) ⇒R=((57)/(14)) ✓

$$\frac{\mathrm{4}\:\mathrm{cos}\:\theta+\mathrm{5}\:\mathrm{cos}\:\theta}{\mathrm{2}}={R} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{2}{R}}{\mathrm{9}} \\ $$$$\left(\mathrm{2}{R}\right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{2}×\mathrm{10}×\mathrm{9}\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\mathrm{4}{R}^{\mathrm{2}} =\mathrm{181}−\mathrm{180}\left(\mathrm{2}×\frac{\mathrm{4}{R}^{\mathrm{2}} }{\mathrm{81}}−\mathrm{1}\right) \\ $$$${R}^{\mathrm{2}} =\frac{\mathrm{3249}}{\mathrm{196}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{57}}{\mathrm{14}}\:\checkmark \\ $$

Commented by cherokeesay last updated on 03/May/24

$${thank}\:{you}\:{master}\:! \\ $$

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