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Question-4971

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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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