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Question-41348

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Question Number 41348 by behi83417@gmail.com last updated on 06/Aug/18
Commented by MJS last updated on 06/Aug/18
trapezoid ABCD a=AB b=BC c=CD d=DA e=AC f=BD which triangles do you mean? △ABC and △ABD?
$$\mathrm{trapezoid}\:{ABCD} \\ $$$${a}={AB}\:\:{b}={BC}\:\:{c}={CD}\:\:{d}={DA} \\ $$$${e}={AC}\:\:{f}={BD} \\ $$$$\mathrm{which}\:\mathrm{triangles}\:\mathrm{do}\:\mathrm{you}\:\mathrm{mean}? \\ $$$$\bigtriangleup{ABC}\:\mathrm{and}\:\bigtriangleup{ABD}? \\ $$
Commented by behi83417@gmail.com last updated on 06/Aug/18
there is 4 triangles.you can assume any 2 triangles with areas m&n.2 opposite or 2 neighbor triangles.question have more than 1 ansewer.
$${there}\:{is}\:\mathrm{4}\:{triangles}.{you}\:{can}\: \\ $$$${assume}\:{any}\:\mathrm{2}\:{triangles}\:{with} \\ $$$${areas}\:{m\&n}.\mathrm{2}\:{opposite}\:{or}\:\mathrm{2}\:{neighbor} \\ $$$${triangles}.{question}\:{have}\:{more}\: \\ $$$${than}\:\mathrm{1}\:{ansewer}. \\ $$
Commented by behi83417@gmail.com last updated on 06/Aug/18
Commented by behi83417@gmail.com last updated on 06/Aug/18
Commented by behi83417@gmail.com last updated on 06/Aug/18
△oAB∼△oDC⇒((oA)/(oC))=((oB)/(oD))=((AB)/(DC))=k (S_(oAB) /S_(oDC) )=k^2 ⇒(m/n)=k^2 ⇒k=(√(m/n)) S_(ABCD) =(1/2)(AB+CD).AP=(1/2)(AB+CD)(oH+oH′)= =(1/2)(AB+CD)(((2m)/(AB))+((2n)/(CD)))= =(1/2)(AB.((2m)/(AB))+AB.((2n)/(CD))+CD.((2m)/(AB))+CD.((2n)/(CD)))= =(1/2)(2m+2n.(√(m/n))+2m.(√(n/m))+2n)= =(m+2(√(mn))+n)=((√m)+(√n))^2 .
$$\bigtriangleup{oAB}\sim\bigtriangleup{oDC}\Rightarrow\frac{{oA}}{{oC}}=\frac{{oB}}{{oD}}=\frac{{AB}}{{DC}}={k} \\ $$$$\frac{{S}_{{oAB}} }{{S}_{{oDC}} }={k}^{\mathrm{2}} \Rightarrow\frac{{m}}{{n}}={k}^{\mathrm{2}} \Rightarrow{k}=\sqrt{\frac{{m}}{{n}}} \\ $$$${S}_{{ABCD}} =\frac{\mathrm{1}}{\mathrm{2}}\left({AB}+{CD}\right).{AP}=\frac{\mathrm{1}}{\mathrm{2}}\left({AB}+{CD}\right)\left({oH}+{oH}'\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({AB}+{CD}\right)\left(\frac{\mathrm{2}{m}}{{AB}}+\frac{\mathrm{2}{n}}{{CD}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({AB}.\frac{\mathrm{2}{m}}{{AB}}+{AB}.\frac{\mathrm{2}{n}}{{CD}}+{CD}.\frac{\mathrm{2}{m}}{{AB}}+{CD}.\frac{\mathrm{2}{n}}{{CD}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{m}+\mathrm{2}{n}.\sqrt{\frac{{m}}{{n}}}+\mathrm{2}{m}.\sqrt{\frac{{n}}{{m}}}+\mathrm{2}{n}\right)= \\ $$$$=\left({m}+\mathrm{2}\sqrt{{mn}}+{n}\right)=\left(\sqrt{{m}}+\sqrt{{n}}\right)^{\mathrm{2}} . \\ $$

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