Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 5354 by Rasheed Soomro last updated on 11/May/16

Commented by Rasheed Soomro last updated on 11/May/16

ABCD and EFGH are squares in above figure.The centres of the arcs are vertices of the external square. If EF=t then AB=?

$$\mathrm{ABCD}\:\mathrm{and}\:\mathrm{EFGH}\:\mathrm{are}\:\mathrm{squares}\:\mathrm{in} \\ $$$$\mathrm{above}\:\mathrm{figure}.\mathrm{The}\:\mathrm{centres}\:\mathrm{of}\:\mathrm{the}\:\:\mathrm{arcs}\: \\ $$$$\mathrm{are}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{the}\:\mathrm{external}\:\mathrm{square}. \\ $$$$\mathrm{If}\:\:\mathrm{EF}=\mathrm{t}\:\mathrm{then}\:\mathrm{AB}=? \\ $$

Commented by Yozzii last updated on 11/May/16

Let sides DA and AB lie on the y and x axes respectively, with A having coordinates (0,0). So let B have coordinates B(r,0) r>0. Since FG=t>0, for EGFH being a square of side t, its sides are parallel to those of □ABCD and its centre coincides with that of □ABCD, then the line x=r/2 reveals a symmetry of the figure. Hence, the x−coordinate of point F is given by x=((r−t)/2). Since F lies on the diagonal AC which coincides with the line y=x, then at F, y=((r−t)/2). So, point F has coordinates (((r−t)/2),((r−t)/2)). Now, the point H is the point F translated t units horizontally and vertically towards the side BC of □ABCD. So, relative to A, H has coordinates (((r−t)/2)+t,((r−t)/2)+t)=(((r+t)/2),((r+t)/2)). But, AH=AB=r since ABH forms a sector of a circle with A as the centre. Therefore, (((r+t)/2)−0)^2 +(((r+t)/2)−0)^2 =r^2 2(((r+t)/2))^2 =r^2 ⇒((r+t)/2)=(r/(√2))⇒r+t=r(√2) r((√2)−1)=t⇒r=(t/((√2)−1))=t(1+(√2)) So, if EF=t⇒AB=t(1+(√2)).

$${Let}\:{sides}\:{DA}\:{and}\:{AB}\:{lie}\:{on}\:{the}\:{y}\:{and}\:{x} \\ $$$${axes}\:{respectively},\:{with}\:{A}\:{having}\: \\ $$$${coordinates}\:\left(\mathrm{0},\mathrm{0}\right).\:{So}\:{let}\:{B}\:{have}\:{coordinates}\:{B}\left({r},\mathrm{0}\right) \\ $$$${r}>\mathrm{0}.\:{Since}\:{FG}={t}>\mathrm{0},\:{for}\:{EGFH}\:{being} \\ $$$${a}\:{square}\:{of}\:{side}\:{t},\:{its}\:{sides}\:{are}\:{parallel} \\ $$$${to}\:{those}\:{of}\:\Box{ABCD}\:{and}\:{its}\:{centre}\: \\ $$$${coincides}\:{with}\:{that}\:{of}\:\Box{ABCD},\:{then} \\ $$$${the}\:{line}\:{x}={r}/\mathrm{2}\:{reveals}\:{a}\:{symmetry} \\ $$$${of}\:{the}\:{figure}.\:{Hence},\:{the}\:{x}−{coordinate} \\ $$$${of}\:{point}\:{F}\:{is}\:{given}\:{by}\:{x}=\frac{{r}−{t}}{\mathrm{2}}.\:{Since}\:{F} \\ $$$${lies}\:{on}\:{the}\:{diagonal}\:{AC}\:{which}\:{coincides} \\ $$$${with}\:{the}\:{line}\:{y}={x},\:{then}\:{at}\:{F},\:{y}=\frac{{r}−{t}}{\mathrm{2}}. \\ $$$${So},\:{point}\:{F}\:{has}\:{coordinates}\:\left(\frac{{r}−{t}}{\mathrm{2}},\frac{{r}−{t}}{\mathrm{2}}\right). \\ $$$${Now},\:{the}\:{point}\:{H}\:{is}\:{the}\:{point}\:{F}\:{translated} \\ $$$${t}\:{units}\:{horizontally}\:{and}\:{vertically}\:{towards} \\ $$$${the}\:{side}\:{BC}\:{of}\:\Box{ABCD}.\:{So},\:{relative}\:{to}\:{A}, \\ $$$${H}\:{has}\:{coordinates}\:\left(\frac{{r}−{t}}{\mathrm{2}}+{t},\frac{{r}−{t}}{\mathrm{2}}+{t}\right)=\left(\frac{{r}+{t}}{\mathrm{2}},\frac{{r}+{t}}{\mathrm{2}}\right). \\ $$$${But},\:{AH}={AB}={r}\:{since}\:{ABH}\:{forms}\:{a}\:{sector}\:{of}\:{a}\:{circle}\:{with}\:{A}\:{as}\:{the}\:{centre}.\:{Therefore}, \\ $$$$\left(\frac{{r}+{t}}{\mathrm{2}}−\mathrm{0}\right)^{\mathrm{2}} +\left(\frac{{r}+{t}}{\mathrm{2}}−\mathrm{0}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\frac{{r}+{t}}{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \Rightarrow\frac{{r}+{t}}{\mathrm{2}}=\frac{{r}}{\sqrt{\mathrm{2}}}\Rightarrow{r}+{t}={r}\sqrt{\mathrm{2}} \\ $$$${r}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)={t}\Rightarrow{r}=\frac{{t}}{\sqrt{\mathrm{2}}−\mathrm{1}}={t}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${So},\:{if}\:{EF}={t}\Rightarrow{AB}={t}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right). \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 12/May/16

Good approach!

$$\mathrm{Good}\:\mathrm{approach}! \\ $$

Answered by Rasheed Soomro last updated on 12/May/16

Non-analytical approach. { ((AF+FH=AB)),((CH+FH=CD=AB)) :} { (),() :}AF=CH Let AF=x and Let AB=T ∴CH=x , BC=CD=DA=T since, EF=t So, FG=GH=HE=t FH=(√(t^2 +t^2 ))=(√2) t AF+FH=AB⇒x+(√2) t=T..........................I AC=AF+FH+CH=2x+(√2) t=(√(T^2 +T^2 )) =(√2) T....II 2x+2(√2) t=2T I×2.........III 2x+(√2) t=(√2) T II×1.........IV (√2) t=(2−(√2))T III −IV T=(((√2) t)/(2−(√2)))×((2+(√2))/(2+(√2)))=t((√2)+1)

$$\mathrm{Non}-\mathrm{analytical}\:\mathrm{approach}. \\ $$$$\begin{cases}{\mathrm{AF}+\mathrm{FH}=\mathrm{AB}}\\{\mathrm{CH}+\mathrm{FH}=\mathrm{CD}=\mathrm{AB}}\end{cases}\begin{cases}{}\\{}\end{cases}\mathrm{AF}=\mathrm{CH} \\ $$$$\mathrm{Let}\:\mathrm{AF}=\mathrm{x}\:\mathrm{and}\:\mathrm{Let}\:\mathrm{AB}=\mathrm{T} \\ $$$$\therefore\mathrm{CH}=\mathrm{x}\:,\:\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{T} \\ $$$$\mathrm{since}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{EF}=\mathrm{t} \\ $$$$\mathrm{So},\:\:\:\:\:\:\:\mathrm{FG}=\mathrm{GH}=\mathrm{HE}=\mathrm{t} \\ $$$$\mathrm{FH}=\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }=\sqrt{\mathrm{2}}\:\mathrm{t} \\ $$$$\mathrm{AF}+\mathrm{FH}=\mathrm{AB}\Rightarrow\mathrm{x}+\sqrt{\mathrm{2}}\:\mathrm{t}=\mathrm{T}..........................\mathrm{I} \\ $$$$\mathrm{AC}=\mathrm{AF}+\mathrm{FH}+\mathrm{CH}=\mathrm{2x}+\sqrt{\mathrm{2}}\:\mathrm{t}=\sqrt{\mathrm{T}^{\mathrm{2}} +\mathrm{T}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}}\:\mathrm{T}....\mathrm{II} \\ $$$$\mathrm{2x}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{t}=\mathrm{2T}\:\:\:\:\:\:\mathrm{I}×\mathrm{2}.........\mathrm{III} \\ $$$$\mathrm{2x}+\sqrt{\mathrm{2}}\:\mathrm{t}=\sqrt{\mathrm{2}}\:\mathrm{T}\:\:\:\:\:\mathrm{II}×\mathrm{1}.........\mathrm{IV} \\ $$$$\sqrt{\mathrm{2}}\:\mathrm{t}=\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\mathrm{T}\:\:\:\:\:\mathrm{III}\:−\mathrm{IV} \\ $$$$\mathrm{T}=\frac{\sqrt{\mathrm{2}}\:\mathrm{t}}{\mathrm{2}−\sqrt{\mathrm{2}}}×\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}=\mathrm{t}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$

Commented by Yozzii last updated on 12/May/16

Nice! My understanding of and confidence in Euclidean geometry still needs some work. The education system I′ve gone through placed emphasis on analytical methods.

$${Nice}!\:{My}\:{understanding}\:{of}\:{and}\:{confidence}\:{in}\:{Euclidean}\:{geometry}\:{still} \\ $$$${needs}\:{some}\:{work}.\:{The}\:{education}\:{system}\:{I}'{ve}\:{gone}\:{through} \\ $$$${placed}\:{emphasis}\:{on}\:{analytical}\:{methods}.\: \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 12/May/16

Commented by Rasheed Soomro last updated on 12/May/16

I have been observing you since your joining to this forum.Your logic is very powerful, so Euclidean geometry is simple for you! ..... Have you qualified the cambridge exam/test?

$$\mathrm{I}\:\mathrm{have}\:\mathrm{been}\:\mathrm{observing}\:\mathrm{you}\:\mathrm{since}\:\mathrm{your}\:\mathrm{joining} \\ $$$$\mathrm{to}\:\mathrm{this}\:\mathrm{forum}.\mathrm{Your}\:\mathrm{logic}\:\mathrm{is}\:\mathrm{very}\:\mathrm{powerful},\:\mathrm{so} \\ $$$$\mathrm{Euclidean}\:\mathrm{geometry}\:\mathrm{is}\:\mathrm{simple}\:\mathrm{for}\:\mathrm{you}! \\ $$$$..... \\ $$$$\:\mathrm{Have}\:\mathrm{you}\:\mathrm{qualified}\:\mathrm{the}\:\mathrm{cambridge}\:\mathrm{exam}/\mathrm{test}? \\ $$

Commented by Yozzii last updated on 13/May/16

I′ve only sat the CIE Further Math A Level exam in 2015, and that doesn′t have an emphasis on Euclidean geometry. It appears a bit in studying mechanics, but at that level rigorous proof writing isn′t required. That′s the past. Now I′m just helping myself grow in mathematical skill, rigour and presentation before I begin undergraduate university studies from September 2016 in mathematics.

$${I}'{ve}\:{only}\:{sat}\:{the}\:{CIE}\:{Further}\:{Math} \\ $$$${A}\:{Level}\:{exam}\:{in}\:\mathrm{2015},\:{and}\:{that}\:{doesn}'{t} \\ $$$${have}\:{an}\:{emphasis}\:{on}\:{Euclidean}\:{geometry}. \\ $$$${It}\:{appears}\:{a}\:{bit}\:{in}\:{studying}\:{mechanics}, \\ $$$${but}\:{at}\:{that}\:{level}\:{rigorous}\:{proof}\:{writing} \\ $$$${isn}'{t}\:{required}.\:{That}'{s}\:{the}\:{past}. \\ $$$${Now}\:{I}'{m}\:{just}\:{helping}\:{myself}\:{grow}\: \\ $$$${in}\:{mathematical}\:{skill},\:{rigour}\:{and}\:{presentation} \\ $$$${before}\:{I}\:{begin}\:{undergraduate}\:{university}\:{studies} \\ $$$${from}\:{September}\:\mathrm{2016}\:{in}\:{mathematics}. \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 13/May/16

Th𝛂nkS ! Actually, I didn′t consider any connection between your exam and Eucledean geometry. I only wanted to be informed about your progress.

$$\mathbb{T}\boldsymbol{\mathrm{h}\alpha\mathrm{n}}\Bbbk\mathcal{S}\:!\:\mathrm{Actually},\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{consider}\:\mathrm{any}\:\mathrm{connection} \\ $$$$\mathrm{between}\:\mathrm{your}\:\mathrm{exam}\:\mathrm{and}\:\mathrm{Eucledean}\:\mathrm{geometry}. \\ $$$$\mathrm{I}\:\mathrm{only}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{be}\:\mathrm{informed}\:\mathrm{about}\:\mathrm{your} \\ $$$$\mathrm{progress}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com