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Question Number 75955 by benjo last updated on 21/Dec/19
help me
$${help}\:{me}\: \\ $$$$ \\ $$
Answered by benjo last updated on 21/Dec/19
Answered by mr W last updated on 21/Dec/19
Commented by mr W last updated on 21/Dec/19
cos θ=((7^2 +9^2 −8^2 )/(2×7×9))=((11)/(21)) r_1 =(7/(2×sin θ))=(7/(2×((8(√5))/(21))))=((147)/(16(√5))) r_2 =(9/(2×sin θ))=(9/(2×((8(√5))/(21))))=((189)/(16(√5))) ∠ACB=2((π/2)−θ)+θ=π−θ AB^2 =r_1 ^2 +r_2 ^2 −2r_1 r_2 cos ∠ACB =(((147)/(16(√5))))^2 +(((189)/(16(√5))))^2 +2×((147)/(16(√5)))×((189)/(16(√5)))×((11)/(21)) =((147^2 +189^2 +2×147×189×((11)/(21)))/(16^2 ×5)) =((21609)/(320)) ⇒AB=((147(√5))/(40)) (1/2)×AB×((CD)/2)=(1/2)×r_1 ×r_2 ×sin ∠ACB ((147(√5))/(40))×((CD)/2)=((147)/(16(√5)))×((189)/(16(√5)))×((8(√5))/(21)) ⇒CD=(9/2)=4.5
$$\mathrm{cos}\:\theta=\frac{\mathrm{7}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}×\mathrm{9}}=\frac{\mathrm{11}}{\mathrm{21}} \\ $$$${r}_{\mathrm{1}} =\frac{\mathrm{7}}{\mathrm{2}×\mathrm{sin}\:\theta}=\frac{\mathrm{7}}{\mathrm{2}×\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{21}}}=\frac{\mathrm{147}}{\mathrm{16}\sqrt{\mathrm{5}}} \\ $$$${r}_{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{2}×\mathrm{sin}\:\theta}=\frac{\mathrm{9}}{\mathrm{2}×\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{21}}}=\frac{\mathrm{189}}{\mathrm{16}\sqrt{\mathrm{5}}} \\ $$$$\angle{ACB}=\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\theta\right)+\theta=\pi−\theta \\ $$$${AB}^{\mathrm{2}} ={r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}{r}_{\mathrm{1}} {r}_{\mathrm{2}} \:\mathrm{cos}\:\angle{ACB} \\ $$$$=\left(\frac{\mathrm{147}}{\mathrm{16}\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{189}}{\mathrm{16}\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{147}}{\mathrm{16}\sqrt{\mathrm{5}}}×\frac{\mathrm{189}}{\mathrm{16}\sqrt{\mathrm{5}}}×\frac{\mathrm{11}}{\mathrm{21}} \\ $$$$=\frac{\mathrm{147}^{\mathrm{2}} +\mathrm{189}^{\mathrm{2}} +\mathrm{2}×\mathrm{147}×\mathrm{189}×\frac{\mathrm{11}}{\mathrm{21}}}{\mathrm{16}^{\mathrm{2}} ×\mathrm{5}} \\ $$$$=\frac{\mathrm{21609}}{\mathrm{320}} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{147}\sqrt{\mathrm{5}}}{\mathrm{40}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{AB}×\frac{{CD}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×{r}_{\mathrm{1}} ×{r}_{\mathrm{2}} ×\mathrm{sin}\:\angle{ACB} \\ $$$$\frac{\mathrm{147}\sqrt{\mathrm{5}}}{\mathrm{40}}×\frac{{CD}}{\mathrm{2}}=\frac{\mathrm{147}}{\mathrm{16}\sqrt{\mathrm{5}}}×\frac{\mathrm{189}}{\mathrm{16}\sqrt{\mathrm{5}}}×\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{21}} \\ $$$$\Rightarrow{CD}=\frac{\mathrm{9}}{\mathrm{2}}=\mathrm{4}.\mathrm{5} \\ $$
Commented by benjo last updated on 21/Dec/19
thanks sir.
$$\mathrm{thanks}\:\mathrm{sir}.\: \\ $$

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