Question-213745

Question Number 213745 by efronzo1 last updated on 15/Nov/24

Answered by mnjuly1970 last updated on 16/Nov/24

(Median theorem ): 7^2 + R^2 = 2((√(14)) )^2 + (((2R)^2 )/2) 49 + R^2 = 28 + 2R^2 21 = R^2 ⇒ R= (√(21)) ■

$$\:\left(\mathrm{M}{edian}\:{theorem}\:\right):\:\:\:\:\:\:\mathrm{7}^{\mathrm{2}} \:+\:{R}^{\mathrm{2}} =\:\mathrm{2}\left(\sqrt{\mathrm{14}}\:\right)^{\mathrm{2}} \:+\:\frac{\left(\mathrm{2}{R}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{49}\:+\:{R}^{\mathrm{2}} \:=\:\mathrm{28}\:+\:\mathrm{2}{R}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{21}\:=\:{R}^{\mathrm{2}} \:\Rightarrow\:{R}=\:\sqrt{\mathrm{21}}\:\:\:\:\:\:\:\blacksquare \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

Answered by golsendro last updated on 15/Nov/24

∠ AOC = α (1) cos α = ((2R^2 −14)/(2R^2 )) (2) cos α = ((5R^2 −49)/(4R^2 )) ⇒ ((5R^2 −49)/2) = 2R^2 −14 R^2 = 21⇒R=(√(21))

$$\:\angle\:\mathrm{AOC}\:=\:\alpha \\ $$$$\:\:\left(\mathrm{1}\right)\:\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{2R}^{\mathrm{2}} −\mathrm{14}}{\mathrm{2R}^{\mathrm{2}} } \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{5R}^{\mathrm{2}} −\mathrm{49}}{\mathrm{4R}^{\mathrm{2}} } \\ $$$$\:\:\Rightarrow\:\frac{\mathrm{5R}^{\mathrm{2}} −\mathrm{49}}{\mathrm{2}}\:=\:\mathrm{2R}^{\mathrm{2}} −\mathrm{14} \\ $$$$\:\:\:\:\:\:\:\mathrm{R}^{\mathrm{2}} \:=\:\mathrm{21}\Rightarrow\mathrm{R}=\sqrt{\mathrm{21}} \\ $$

Answered by mr W last updated on 15/Nov/24

4×((√(14)))^2 =2×7^2 +2×R^2 −(2R)^2 ⇒R=(√(7^2 −2×14))=(√(21))

$$\mathrm{4}×\left(\sqrt{\mathrm{14}}\right)^{\mathrm{2}} =\mathrm{2}×\mathrm{7}^{\mathrm{2}} +\mathrm{2}×{R}^{\mathrm{2}} −\left(\mathrm{2}{R}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\sqrt{\mathrm{7}^{\mathrm{2}} −\mathrm{2}×\mathrm{14}}=\sqrt{\mathrm{21}} \\ $$

Answered by A5T last updated on 16/Nov/24

R^2 +14−2R(√(14))cosACO=R^2 ⇒cosACO=((14)/(2R(√(14)))) R^2 +14+2R(√(14))cosACO=49⇒cosACO=((35−R^2 )/(2R(√(14)))) ⇒35−R^2 =14⇒R=(√(21))

$${R}^{\mathrm{2}} +\mathrm{14}−\mathrm{2}{R}\sqrt{\mathrm{14}}{cosACO}={R}^{\mathrm{2}} \Rightarrow{cosACO}=\frac{\mathrm{14}}{\mathrm{2}{R}\sqrt{\mathrm{14}}} \\ $$$${R}^{\mathrm{2}} +\mathrm{14}+\mathrm{2}{R}\sqrt{\mathrm{14}}{cosACO}=\mathrm{49}\Rightarrow{cosACO}=\frac{\mathrm{35}−{R}^{\mathrm{2}} }{\mathrm{2}{R}\sqrt{\mathrm{14}}} \\ $$$$\Rightarrow\mathrm{35}−{R}^{\mathrm{2}} =\mathrm{14}\Rightarrow{R}=\sqrt{\mathrm{21}} \\ $$

Answered by A5T last updated on 16/Nov/24

R^2 ×2R+((√(14)))^2 (2R)=R^2 ×R+7^2 ×R ⇒2R^2 +28=R^2 +49⇒R=(√(21))

$${R}^{\mathrm{2}} ×\mathrm{2}{R}+\left(\sqrt{\mathrm{14}}\right)^{\mathrm{2}} \left(\mathrm{2}{R}\right)={R}^{\mathrm{2}} ×{R}+\mathrm{7}^{\mathrm{2}} ×{R} \\ $$$$\Rightarrow\mathrm{2}{R}^{\mathrm{2}} +\mathrm{28}={R}^{\mathrm{2}} +\mathrm{49}\Rightarrow{R}=\sqrt{\mathrm{21}} \\ $$

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