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Show-that-in-a-30-60-90-triangle-the-altitude-on-the-hypotaneuse-divides-the-hypotaneuse-into-segments-whose-length-has-the-ratio-1-3-without-using-trigonometry-

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Question Number 60043 by Kunal12588 last updated on 17/May/19
Show that in a 30°−60°−90° triangle the altitude on the hypotaneuse divides the hypotaneuse into segments whose length has the ratio 1/3. without using trigonometry.
$${Show}\:{that}\:{in}\:{a}\:\mathrm{30}°−\mathrm{60}°−\mathrm{90}°\:{triangle}\:{the}\: \\ $$$${altitude}\:{on}\:{the}\:{hypotaneuse}\:{divides}\:{the}\: \\ $$$${hypotaneuse}\:{into}\:{segments}\:{whose}\:{length} \\ $$$${has}\:{the}\:{ratio}\:\mathrm{1}/\mathrm{3}. \\ $$$${without}\:{using}\:{trigonometry}. \\ $$
Answered by tanmay last updated on 17/May/19
The eqn of hypotaneous (x/a)+(y/b)=1 bx+ay=ab y=((−bx)/a)+b →slopse m_1 =−(b/a) eq of st line ⊥to hypotaneous and passing through origin(0,0) y=m_2 x m_1 ×m_2 =−1 m_2 =((−1)/((−b)/a))=(a/b) y=(a/b)x solve y=(a/b)x and (x/a)+(y/b)=1 to get point C which devide AB in (1/3) ratio (x/b)=(y/a)=k→x=bk y=ak ((bk)/a)+((ak)/b)=1→k(a^2 +b^2 )=ab k=((ab)/(a^2 +b^2 )) x=((ab^2 )/(a^2 +b^2 )) y=((a^2 b)/(a^2 +b^2 )) A(0,b) B(a,0) O(0,0) and C(((ab^2 )/(a^2 +b^2 )),((a^2 b)/(a^2 +b^2 ))) let C devides AB (m/n) ratio ((ab^2 )/(a^2 +b^2 ))=((m×0+n×a)/(m+n)) mab^2 +nab^2 =na^3 +nab^2 (m/n)=(a^3 /(ab^2 ))=(a^2 /b^2 ) now slope of AB is m_1 =((−b)/a) let AB makes angle θ with OB so slope(m_1 )=((−b)/a)=tan(180^o −60^o )=−(√3) (b/a)=(√3) (m/n)=(a^2 /b^2 )=(1/3) proved ■■ if we take θ=30^o slope=((−b)/a)=tan(180^o −30^o )=−(1/( (√3))) (b/a)=(1/( (√3))) then (m/n)=(a^2 /b^2 )=3 so rstion (3/1) also fulfill... note Triangle OAB OA⊥OB znd AB hypotaneous
$${The}\:{eqn}\:{of}\:{hypotaneous} \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$${bx}+{ay}={ab} \\ $$$${y}=\frac{−{bx}}{{a}}+{b}\:\rightarrow{slopse}\:{m}_{\mathrm{1}} =−\frac{{b}}{{a}} \\ $$$${eq}\:{of}\:{st}\:{line}\:\bot{to}\:{hypotaneous}\:{and}\:{passing} \\ $$$${through}\:{origin}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${y}={m}_{\mathrm{2}} {x}\:\:\:\:{m}_{\mathrm{1}} ×{m}_{\mathrm{2}} =−\mathrm{1} \\ $$$${m}_{\mathrm{2}} =\frac{−\mathrm{1}}{\frac{−{b}}{{a}}}=\frac{{a}}{{b}}\:\:\: \\ $$$${y}=\frac{{a}}{{b}}{x} \\ $$$${solve}\:{y}=\frac{{a}}{{b}}{x}\:{and}\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$${to}\:{get}\:{point}\:{C}\:{which}\:{devide}\:{AB}\:\:\:{in}\:\frac{\mathrm{1}}{\mathrm{3}}\:\:{ratio} \\ $$$$\frac{{x}}{{b}}=\frac{{y}}{{a}}={k}\rightarrow{x}={bk}\:\:\:{y}={ak}\:\: \\ $$$$\frac{{bk}}{{a}}+\frac{{ak}}{{b}}=\mathrm{1}\rightarrow{k}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)={ab}\:\:{k}=\frac{{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${x}=\frac{{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:\:{y}=\frac{{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${A}\left(\mathrm{0},{b}\right)\:\:\:{B}\left({a},\mathrm{0}\right)\:\:\:{O}\left(\mathrm{0},\mathrm{0}\right)\:\:\:{and}\:\boldsymbol{{C}}\left(\frac{{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} },\frac{{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right) \\ $$$${let}\:{C}\:{devides}\:{AB}\:\:\:\:\frac{{m}}{{n}}\:{ratio} \\ $$$$\frac{{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\frac{{m}×\mathrm{0}+{n}×{a}}{{m}+{n}} \\ $$$${mab}^{\mathrm{2}} +{nab}^{\mathrm{2}} ={na}^{\mathrm{3}} +{nab}^{\mathrm{2}} \\ $$$$\frac{{m}}{{n}}=\frac{{a}^{\mathrm{3}} }{{ab}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${now}\:{slope}\:{of}\:{AB}\:{is}\:{m}_{\mathrm{1}} =\frac{−{b}}{{a}} \\ $$$${let}\:{AB}\:{makes}\:{angle}\:\theta\:{with}\:{OB} \\ $$$${so}\:\:{slope}\left({m}_{\mathrm{1}} \right)=\frac{−{b}}{{a}}={tan}\left(\mathrm{180}^{{o}} −\mathrm{60}^{{o}} \right)=−\sqrt{\mathrm{3}}\: \\ $$$$\frac{{b}}{{a}}=\sqrt{\mathrm{3}}\: \\ $$$$\frac{{m}}{{n}}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{3}}\:\:{proved} \\ $$$$\blacksquare\blacksquare\:{if}\:{we}\:{take}\:\theta=\mathrm{30}^{{o}} \\ $$$${slope}=\frac{−{b}}{{a}}={tan}\left(\mathrm{180}^{{o}} −\mathrm{30}^{{o}} \right)=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${then}\:\frac{{m}}{{n}}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{3}\: \\ $$$${so}\:{rstion}\:\frac{\mathrm{3}}{\mathrm{1}}\:{also}\:{fulfill}… \\ $$$$ \\ $$$${note}\:{Triangle}\:{OAB} \\ $$$${OA}\bot{OB}\:\:\:{znd}\:{AB}\:{hypotaneous} \\ $$$$ \\ $$
Commented by Kunal12588 last updated on 17/May/19
sir you still used trigonometry (tangent)
$${sir}\:{you}\:{still}\:{used}\:{trigonometry}\:\left({tangent}\right) \\ $$
Commented by Kunal12588 last updated on 17/May/19
but i like your way sir (pls let it be here)
$${but}\:{i}\:{like}\:{your}\:{way}\:{sir}\:\left({pls}\:{let}\:{it}\:{be}\:{here}\right) \\ $$
Answered by mr W last updated on 17/May/19
Commented by mr W last updated on 17/May/19
let AB=1, AF=x ⇒BC=CE=1 ⇒CD=x ⇒FE=ED=(√(x^2 −1)) ⇒AD=2+x AD^2 =AF^2 +FD^2 ⇒(2+x)^2 =x^2 +(2(√(x^2 −1)))^2 ⇒4+4x+x^2 =x^2 +4x^2 −4 ⇒x^2 −x−2=0 ⇒(x+1)(x−2)=0 ⇒x=2 ⇒BD=1+2=3 ⇒AB/BD=1/3 ⇒proved
$${let}\:{AB}=\mathrm{1},\:{AF}={x} \\ $$$$\Rightarrow{BC}={CE}=\mathrm{1} \\ $$$$\Rightarrow{CD}={x} \\ $$$$\Rightarrow{FE}={ED}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{AD}=\mathrm{2}+{x} \\ $$$${AD}^{\mathrm{2}} ={AF}^{\mathrm{2}} +{FD}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}+{x}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}+\mathrm{4}{x}+{x}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$$\Rightarrow{BD}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\Rightarrow{AB}/{BD}=\mathrm{1}/\mathrm{3}\:\Rightarrow{proved} \\ $$
Commented by Kunal12588 last updated on 17/May/19
sir i tried pls check △AFC is equilateral ⇒FC=AC=AB+BC=2 FC=DC ⇒AC=DC ⇒x=2 ((AB)/(BD))=(1/(x+1))=(1/(2+1))=(1/3)
$${sir}\:{i}\:{tried}\:{pls}\:{check} \\ $$$$\bigtriangleup{AFC}\:{is}\:{equilateral}\: \\ $$$$\Rightarrow{FC}={AC}={AB}+{BC}=\mathrm{2} \\ $$$${FC}={DC} \\ $$$$\Rightarrow{AC}={DC} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$$\frac{{AB}}{{BD}}=\frac{\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by mr W last updated on 17/May/19
nice path sir!
$${nice}\:{path}\:{sir}! \\ $$

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