How-do-I-solve-this-please-Show-that-if-the-sides-of-a-right-triangle-are-in-an-arithmetic-sequence-then-their-ratio-is-3-4-5-

Question Number 199986 by Yakubu last updated on 11/Nov/23

$$ \\ $$How do I solve this please?
Show that if the sides of a right triangle are in an arithmetic sequence, then their ratio is 3:4:5.

Answered by Frix last updated on 12/Nov/23

a^2 +(a+k)^2 =(a+2k)^2 with a, k >0 a^2 −2ka−3k^2 =0 (a+k)(a−3k)=0 k=−a impossible k=(a/3) All triangles with sides a, ((4a)/3), ((5a)/3) are rectangular and their sides are in an arithmetic sequence and their ratio is 3:4:5

$${a}^{\mathrm{2}} +\left({a}+{k}\right)^{\mathrm{2}} =\left({a}+\mathrm{2}{k}\right)^{\mathrm{2}} \:\mathrm{with}\:{a},\:{k}\:>\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{ka}−\mathrm{3}{k}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}+{k}\right)\left({a}−\mathrm{3}{k}\right)=\mathrm{0} \\ $$$${k}=−{a}\:\mathrm{impossible} \\ $$$${k}=\frac{{a}}{\mathrm{3}} \\ $$$$\mathrm{All}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sides}\:{a},\:\frac{\mathrm{4}{a}}{\mathrm{3}},\:\frac{\mathrm{5}{a}}{\mathrm{3}}\:\mathrm{are} \\ $$$$\mathrm{rectangular}\:\mathrm{and}\:\mathrm{their}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{in}\:\mathrm{an} \\ $$$$\mathrm{arithmetic}\:\mathrm{sequence}\:\mathrm{and}\:\mathrm{their}\:\mathrm{ratio}\:\mathrm{is} \\ $$$$\mathrm{3}:\mathrm{4}:\mathrm{5} \\ $$

Commented by Yakubu last updated on 12/Nov/23

$${thank}\:{you}\:{sir}. \\ $$

Answered by Rasheed.Sindhi last updated on 12/Nov/23

AnOther way: Let the required traingle has sides in ratio l:m:n We′ll prove that l:m:n=3:4:5 la,ma,na are sides { (((la)^2 +(ma)^2 =(na)^2 ⇒l^2 +m^2 =n^2 )),((ma=((la+na)/2)⇒m=((l+n)/2))) :} l^2 +(((l+n)/2))^2 =n^2 4l^2 +l^2 +2ln+n^2 =4n^2 5l^2 +2ln=3n^2 ((5l)/(2n))+1=((3n)/(2l)) (5/2)t−(3/(2t))+1=0 5t^2 +2t−3=0 (t+1)(5t−3)=0 t=(l/n)=−1(absurd ∵ l,n∈Z^+ ∣ t=(3/5)⇒(l/n)=(3/5)⇒n=((5l)/3) m=((l+n)/2)⇒2m=l+((5l)/3) ⇒6m=8l⇒(l/m)=(3/4) (l/n)=(3/5) ∧ (l/m)=(3/4)⇒ determinant (((l:m:n=3:4:5))) proved

$$\mathrm{AnOther}\:\mathrm{way}: \\ $$$${Let}\:{the}\:{required}\:{traingle}\:{has} \\ $$$${sides}\:{in}\:{ratio}\:\:{l}:{m}:{n} \\ $$$${We}'{ll}\:{prove}\:{that}\:{l}:{m}:{n}=\mathrm{3}:\mathrm{4}:\mathrm{5} \\ $$$${la},{ma},{na}\:{are}\:{sides} \\ $$$$\begin{cases}{\left({la}\right)^{\mathrm{2}} +\left({ma}\right)^{\mathrm{2}} =\left({na}\right)^{\mathrm{2}} \Rightarrow{l}^{\mathrm{2}} +{m}^{\mathrm{2}} ={n}^{\mathrm{2}} }\\{{ma}=\frac{{la}+{na}}{\mathrm{2}}\Rightarrow{m}=\frac{{l}+{n}}{\mathrm{2}}}\end{cases} \\ $$$${l}^{\mathrm{2}} +\left(\frac{{l}+{n}}{\mathrm{2}}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} \\ $$$$\mathrm{4}{l}^{\mathrm{2}} +{l}^{\mathrm{2}} +\mathrm{2}{ln}+{n}^{\mathrm{2}} =\mathrm{4}{n}^{\mathrm{2}} \\ $$$$\mathrm{5}{l}^{\mathrm{2}} +\mathrm{2}{ln}=\mathrm{3}{n}^{\mathrm{2}} \\ $$$$\frac{\mathrm{5}{l}}{\mathrm{2}{n}}+\mathrm{1}=\frac{\mathrm{3}{n}}{\mathrm{2}{l}} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}{t}−\frac{\mathrm{3}}{\mathrm{2}{t}}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{5}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{3}=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left(\mathrm{5}{t}−\mathrm{3}\right)=\mathrm{0} \\ $$$${t}=\frac{{l}}{{n}}=−\mathrm{1}\left({absurd}\:\because\:{l},{n}\in\mathbb{Z}^{+} \mid\right. \\ $$$$\:{t}=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow\frac{{l}}{{n}}=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow{n}=\frac{\mathrm{5}{l}}{\mathrm{3}} \\ $$$${m}=\frac{{l}+{n}}{\mathrm{2}}\Rightarrow\mathrm{2}{m}={l}+\frac{\mathrm{5}{l}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{6}{m}=\mathrm{8}{l}\Rightarrow\frac{{l}}{{m}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{{l}}{{n}}=\frac{\mathrm{3}}{\mathrm{5}}\:\wedge\:\frac{{l}}{{m}}=\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow\begin{array}{|c|}{{l}:{m}:{n}=\mathrm{3}:\mathrm{4}:\mathrm{5}}\\\hline\end{array} \\ $$$${proved} \\ $$

Commented by Yakubu last updated on 12/Nov/23

$${thank}\:{you}\:{sir} \\ $$

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