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Question-4971




Question Number 4971 by Rasheed Soomro last updated on 28/Mar/16
Commented by Rasheed Soomro last updated on 28/Mar/16
Write process also.
$${Write}\:{process}\:{also}. \\ $$
Commented by Rasheed Soomro last updated on 29/Mar/16
How?
$${How}? \\ $$
Commented by Yozzii last updated on 29/Mar/16
one−quarter the area of the square
$${one}−{quarter}\:{the}\:{area}\:{of}\:{the}\:{square} \\ $$
Commented by Yozzii last updated on 29/Mar/16
The lines from the midpoints on the sides of the square  trissect the hypotenuse of the smaller triangle. I was  trying to upload my answer as pics  but it failed to upload.
$${The}\:{lines}\:{from}\:{the}\:{midpoints}\:{on}\:{the}\:{sides}\:{of}\:{the}\:{square} \\ $$$${trissect}\:{the}\:{hypotenuse}\:{of}\:{the}\:{smaller}\:{triangle}.\:{I}\:{was} \\ $$$${trying}\:{to}\:{upload}\:{my}\:{answer}\:{as}\:{pics} \\ $$$${but}\:{it}\:{failed}\:{to}\:{upload}. \\ $$
Commented by Yozzii last updated on 29/Mar/16
Commented by Yozzii last updated on 30/Mar/16
The inner triangles are reflections of  each other in the diagonal. The diagonal  of the square is also trissected. So  ST  one third AC. ST,AC,RQ and NJ  are parallel. So, the area you want  is made up of areas of two congruent  trapezia. △ROQ is isosceles so   the line from O that falls on RQ   normally bissects the line RQ.  Using this, the length OM is found  to be ((a(√2))/4). Likewise, BF=((a(√2))/4) since  △ROQ≊△BNJ. So, MF=OB−(BF+OM)  MF=a(√2)−((2a(√2))/4)=((a(√2))/2).  ((MF)/2) is the ′height′ of each trapezium.  You see in my diagram that LK=(1/3)RQ=((a(√2))/6)  and ST=((a(√2))/3).   ∴ A_t =(1/2)(LK+ST)((MF)/2)+(1/2)(UV+ST)((MF)/2)  A_t =((MF)/4)(LK+UV+2ST)  A_t =((a(√2)/2)/4)(2LK+2ST)  A_t =((a(√2))/4)(((a(√2))/6)+((a(√2))/3))  A_t =((2a^2 )/4)((9/(18)))=(a^2 /4).
$${The}\:{inner}\:{triangles}\:{are}\:{reflections}\:{of} \\ $$$${each}\:{other}\:{in}\:{the}\:{diagonal}.\:{The}\:{diagonal} \\ $$$${of}\:{the}\:{square}\:{is}\:{also}\:{trissected}.\:{So} \\ $$$${ST}\:\:{one}\:{third}\:{AC}.\:{ST},{AC},{RQ}\:{and}\:{NJ} \\ $$$${are}\:{parallel}.\:{So},\:{the}\:{area}\:{you}\:{want} \\ $$$${is}\:{made}\:{up}\:{of}\:{areas}\:{of}\:{two}\:{congruent} \\ $$$${trapezia}.\:\bigtriangleup{ROQ}\:{is}\:{isosceles}\:{so}\: \\ $$$${the}\:{line}\:{from}\:{O}\:{that}\:{falls}\:{on}\:{RQ}\: \\ $$$${normally}\:{bissects}\:{the}\:{line}\:{RQ}. \\ $$$${Using}\:{this},\:{the}\:{length}\:{OM}\:{is}\:{found} \\ $$$${to}\:{be}\:\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{4}}.\:{Likewise},\:{BF}=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{4}}\:{since} \\ $$$$\bigtriangleup{ROQ}\approxeq\bigtriangleup{BNJ}.\:{So},\:{MF}={OB}−\left({BF}+{OM}\right) \\ $$$${MF}={a}\sqrt{\mathrm{2}}−\frac{\mathrm{2}{a}\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{2}}. \\ $$$$\frac{{MF}}{\mathrm{2}}\:{is}\:{the}\:'{height}'\:{of}\:{each}\:{trapezium}. \\ $$$${You}\:{see}\:{in}\:{my}\:{diagram}\:{that}\:{LK}=\frac{\mathrm{1}}{\mathrm{3}}{RQ}=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{6}} \\ $$$${and}\:{ST}=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{3}}.\: \\ $$$$\therefore\:{A}_{{t}} =\frac{\mathrm{1}}{\mathrm{2}}\left({LK}+{ST}\right)\frac{{MF}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left({UV}+{ST}\right)\frac{{MF}}{\mathrm{2}} \\ $$$${A}_{{t}} =\frac{{MF}}{\mathrm{4}}\left({LK}+{UV}+\mathrm{2}{ST}\right) \\ $$$${A}_{{t}} =\frac{{a}\sqrt{\mathrm{2}}/\mathrm{2}}{\mathrm{4}}\left(\mathrm{2}{LK}+\mathrm{2}{ST}\right) \\ $$$${A}_{{t}} =\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{6}}+\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$${A}_{{t}} =\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{\mathrm{9}}{\mathrm{18}}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}. \\ $$
Commented by Yozzii last updated on 30/Mar/16
The original solution I wanted to share  is a bit lengthy because with only assuming  that the figure is a square and that  the large inner triangles have vertices  as 2 side mid−points and 1 vertex of the  square, I had to prove everything  else, such as parallelism and trissection.  It wasn′t specified if the figure was a rectangle  or square or if those vertices are   really midpoints of the sides of the   figure. One could attempt the   general case where the vertices of   the triangle come from one rectangle  vertex and two points that lie anywhere  on two sides of the rectangle that do not contain the  rectangle vertex.
$${The}\:{original}\:{solution}\:{I}\:{wanted}\:{to}\:{share} \\ $$$${is}\:{a}\:{bit}\:{lengthy}\:{because}\:{with}\:{only}\:{assuming} \\ $$$${that}\:{the}\:{figure}\:{is}\:{a}\:{square}\:{and}\:{that} \\ $$$${the}\:{large}\:{inner}\:{triangles}\:{have}\:{vertices} \\ $$$${as}\:\mathrm{2}\:{side}\:{mid}−{points}\:{and}\:\mathrm{1}\:{vertex}\:{of}\:{the} \\ $$$${square},\:{I}\:{had}\:{to}\:{prove}\:{everything} \\ $$$${else},\:{such}\:{as}\:{parallelism}\:{and}\:{trissection}. \\ $$$${It}\:{wasn}'{t}\:{specified}\:{if}\:{the}\:{figure}\:{was}\:{a}\:{rectangle} \\ $$$${or}\:{square}\:{or}\:{if}\:{those}\:{vertices}\:{are}\: \\ $$$${really}\:{midpoints}\:{of}\:{the}\:{sides}\:{of}\:{the}\: \\ $$$${figure}.\:{One}\:{could}\:{attempt}\:{the}\: \\ $$$${general}\:{case}\:{where}\:{the}\:{vertices}\:{of}\: \\ $$$${the}\:{triangle}\:{come}\:{from}\:{one}\:{rectangle} \\ $$$${vertex}\:{and}\:{two}\:{points}\:{that}\:{lie}\:{anywhere} \\ $$$${on}\:{two}\:{sides}\:{of}\:{the}\:{rectangle}\:{that}\:{do}\:{not}\:{contain}\:{the} \\ $$$${rectangle}\:{vertex}. \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 30/Mar/16
Commented by Rasheed Soomro last updated on 30/Mar/16
All small triangles have equal  area. Some triangles are congruent  also. There are some  triangles,which are  not congruent but despite this  they have equal  area, because they have equal bases  and altitudes.For example:  △AIQ≇△IQM  But these triangles have equal bases  AQ=QM and common altitude IV  and hence equal area.  Now number of triangles in whole  is 24,  Number of triangles in shaded area  is 6.  ∴ The shaded area is (6/(24))=(1/4) part of the  whole.
$${All}\:{small}\:{triangles}\:{have}\:{equal} \\ $$$${area}.\:{Some}\:{triangles}\:{are}\:{congruent} \\ $$$${also}.\:{There}\:{are}\:{some}\:\:{triangles},{which}\:{are} \\ $$$${not}\:{congruent}\:{but}\:{despite}\:{this}\:\:{they}\:{have}\:{equal} \\ $$$${area},\:{because}\:{they}\:{have}\:{equal}\:{bases} \\ $$$${and}\:{altitudes}.{For}\:{example}: \\ $$$$\bigtriangleup{AIQ}\ncong\bigtriangleup{IQM} \\ $$$${But}\:{these}\:{triangles}\:{have}\:{equal}\:{bases} \\ $$$${AQ}={QM}\:{and}\:{common}\:{altitude}\:{IV} \\ $$$${and}\:{hence}\:{equal}\:{area}. \\ $$$${Now}\:{number}\:{of}\:{triangles}\:{in}\:{whole} \\ $$$${is}\:\mathrm{24}, \\ $$$${Number}\:{of}\:{triangles}\:{in}\:{shaded}\:{area} \\ $$$${is}\:\mathrm{6}. \\ $$$$\therefore\:{The}\:{shaded}\:{area}\:{is}\:\frac{\mathrm{6}}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{4}}\:{part}\:{of}\:{the} \\ $$$${whole}. \\ $$$$ \\ $$
Commented by Yozzii last updated on 30/Mar/16
Nice.
$${Nice}. \\ $$
Commented by FilupSmith last updated on 31/Mar/16
Amazing and simple to understand!
$${A}\mathrm{mazing}\:\mathrm{and}\:\mathrm{simple}\:\mathrm{to}\:\mathrm{understand}! \\ $$
Answered by Rasheed Soomro last updated on 31/Mar/16
Commented by Rasheed Soomro last updated on 31/Mar/16
An other answer:  H,I,J & K are midpoints of the  sides of the □ABCD and it can  be shown that:  ■HIJK=(1/2) of ■ABCD (■ for area        (I)  of square)  Now,  □HIJK contains  24 small triangles  and  polygon PQRSTU contains 12 small  triangles.  So,  Area PQRSTU=((12)/(24))=(1/2) of ■HIJK                  (II)  From (I) & (II)  Area PQRSTU=(1/2)×(1/2)=(1/4) of ■ABCD
$${An}\:{other}\:{answer}: \\ $$$$\mathrm{H},\mathrm{I},\mathrm{J}\:\&\:\mathrm{K}\:{are}\:{midpoints}\:{of}\:{the} \\ $$$${sides}\:{of}\:{the}\:\Box\mathrm{ABCD}\:{and}\:{it}\:{can} \\ $$$${be}\:{shown}\:{that}: \\ $$$$\blacksquare\mathrm{HIJK}=\frac{\mathrm{1}}{\mathrm{2}}\:{of}\:\blacksquare\mathrm{ABCD}\:\left(\blacksquare\:{for}\:{area}\:\:\:\:\:\:\:\:\left(\mathrm{I}\right)\right. \\ $$$$\left.{of}\:{square}\right) \\ $$$${Now}, \\ $$$$\Box\mathrm{HIJK}\:{contains}\:\:\mathrm{24}\:{small}\:{triangles} \\ $$$${and} \\ $$$${polygon}\:\mathrm{PQRSTU}\:{contains}\:\mathrm{12}\:{small} \\ $$$${triangles}. \\ $$$${So}, \\ $$$${Area}\:\mathrm{PQRSTU}=\frac{\mathrm{12}}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{2}}\:{of}\:\blacksquare\mathrm{HIJK}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{II}\right) \\ $$$${From}\:\left(\mathrm{I}\right)\:\&\:\left(\mathrm{II}\right) \\ $$$${Area}\:\mathrm{PQRSTU}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\:{of}\:\blacksquare\mathrm{ABCD} \\ $$

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