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Question-186577




Question Number 186577 by Mingma last updated on 06/Feb/23
Answered by mr W last updated on 06/Feb/23
Commented by mr W last updated on 06/Feb/23
(5/(15))=(3/9) ⇒x//c  (x/c)=((3+9)/3)=4  a^2 +16b^2 =15^2   16a^2 +b^2 =9^2   ⇒a^2 =((16×9^2 −15^2 )/(16^2 −1))=4.2  ⇒b^2 =((16×15^2 −9^2 )/(16^2 −1))=13.8  c=(√(a^2 +b^2 ))=(√(18))=3(√2)  ⇒x=4c=12(√2) ✓
$$\frac{\mathrm{5}}{\mathrm{15}}=\frac{\mathrm{3}}{\mathrm{9}}\:\Rightarrow{x}//{c} \\ $$$$\frac{{x}}{{c}}=\frac{\mathrm{3}+\mathrm{9}}{\mathrm{3}}=\mathrm{4} \\ $$$${a}^{\mathrm{2}} +\mathrm{16}{b}^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} \\ $$$$\mathrm{16}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{16}×\mathrm{9}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{2}} −\mathrm{1}}=\mathrm{4}.\mathrm{2} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\frac{\mathrm{16}×\mathrm{15}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{2}} −\mathrm{1}}=\mathrm{13}.\mathrm{8} \\ $$$${c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\sqrt{\mathrm{18}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{4}{c}=\mathrm{12}\sqrt{\mathrm{2}}\:\checkmark \\ $$

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