Menu Close

Question-189114




Question Number 189114 by normans last updated on 12/Mar/23
Answered by mr W last updated on 12/Mar/23
Commented by mr W last updated on 12/Mar/23
c=(√(a^2 +(b−a)^2 ))  (((b−a)a)/c)+c+((ab)/c)=13  ⇒a^2 +b^2 =13c   ...(i)  (((b−a)b)/c)+((ab)/c)=9  ⇒b^2 =9c   ...(ii)  a^2 =((4b^2 )/9) ⇒a=((2b)/3)  b^2 =9(√(a^2 +(b−a)^2 ))  ⇒b=3(√5) ⇒a=2(√5)  ⇒b^2 =45, a^2 =20
$${c}=\sqrt{{a}^{\mathrm{2}} +\left({b}−{a}\right)^{\mathrm{2}} } \\ $$$$\frac{\left({b}−{a}\right){a}}{{c}}+{c}+\frac{{ab}}{{c}}=\mathrm{13} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{13}{c}\:\:\:…\left({i}\right) \\ $$$$\frac{\left({b}−{a}\right){b}}{{c}}+\frac{{ab}}{{c}}=\mathrm{9} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\mathrm{9}{c}\:\:\:…\left({ii}\right) \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{9}}\:\Rightarrow{a}=\frac{\mathrm{2}{b}}{\mathrm{3}} \\ $$$${b}^{\mathrm{2}} =\mathrm{9}\sqrt{{a}^{\mathrm{2}} +\left({b}−{a}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{b}=\mathrm{3}\sqrt{\mathrm{5}}\:\Rightarrow{a}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\mathrm{45},\:{a}^{\mathrm{2}} =\mathrm{20} \\ $$
Answered by HeferH last updated on 12/Mar/23
Commented by HeferH last updated on 12/Mar/23
∗ (x/4) = (b/a); x = ((4b)/a)  ∗ (b/(b + a)) = ((9−((4b)/a))/5)   5b = 9(b+a) − ((4b(b+a))/a)   5ab = 9ab + 9a^2  −4b^2  −4ab   4b^2  = 9a^2    (b/a) = (√(9/4)) = (3/2)   ⇒ x = 6   ∗ now:   (5−3)^2  + 4^2  = a^2  = 20u^2    (9 − 6)^2  + 6^2  = b^2  = 45u^2
$$\ast\:\frac{{x}}{\mathrm{4}}\:=\:\frac{{b}}{{a}};\:{x}\:=\:\frac{\mathrm{4}{b}}{{a}} \\ $$$$\ast\:\frac{{b}}{{b}\:+\:{a}}\:=\:\frac{\mathrm{9}−\frac{\mathrm{4}{b}}{{a}}}{\mathrm{5}} \\ $$$$\:\mathrm{5}{b}\:=\:\mathrm{9}\left({b}+{a}\right)\:−\:\frac{\mathrm{4}{b}\left({b}+{a}\right)}{{a}} \\ $$$$\:\mathrm{5}{ab}\:=\:\mathrm{9}{ab}\:+\:\mathrm{9}{a}^{\mathrm{2}} \:−\mathrm{4}{b}^{\mathrm{2}} \:−\mathrm{4}{ab} \\ $$$$\:\mathrm{4}{b}^{\mathrm{2}} \:=\:\mathrm{9}{a}^{\mathrm{2}} \\ $$$$\:\frac{{b}}{{a}}\:=\:\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\Rightarrow\:{x}\:=\:\mathrm{6} \\ $$$$\:\ast\:\mathrm{now}: \\ $$$$\:\left(\mathrm{5}−\mathrm{3}\right)^{\mathrm{2}} \:+\:\mathrm{4}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:=\:\mathrm{20}{u}^{\mathrm{2}} \\ $$$$\:\left(\mathrm{9}\:−\:\mathrm{6}\right)^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \:=\:\mathrm{45}{u}^{\mathrm{2}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *