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Question-15318




Question Number 15318 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17
given: ABCD,rectangle.connect   each vertex to the middle point of  one other sides as shown.  1)show that MNPQ,is a parallelogram  and its area is equail to: (1/5)area of ABCD.  2)S_(SP^Δ C) =S_(BN^Δ T) =(1/(20)).S_(ABCD) .
given:ABCD,rectangle.connecteachvertextothemiddlepointofoneothersidesasshown.1)showthatMNPQ,isaparallelogramanditsareaisequailto:15areaofABCD.2)SSPCΔ=SBNTΔ=120.SABCD.
Commented by mrW1 last updated on 09/Jun/17
Commented by mrW1 last updated on 09/Jun/17
S_1 =S_2 =S_3 =S_4 =S_5   S_1 +S_2 +S_3 +S_4 +S_5 =S_(ABCD)   ⇒S_1 =(1/5)S_(ABCD)     S_(ΔSPC) =(1/4)S_4 =(1/4)×(1/5)S_(ABCD) =(1/(20))S_(ABCD)   S_(ΔBNT) =(1/4)S_3 =(1/4)×(1/5)S_(ABCD) =(1/(20))S_(ABCD)
S1=S2=S3=S4=S5S1+S2+S3+S4+S5=SABCDS1=15SABCDSΔSPC=14S4=14×15SABCD=120SABCDSΔBNT=14S3=14×15SABCD=120SABCD
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17
thank you master.it is simple and perfect.
thankyoumaster.itissimpleandperfect.
Commented by mrW1 last updated on 09/Jun/17
what if not the middle point but the  1/3−point is connected with the  vertex?
whatifnotthemiddlepointbutthe1/3pointisconnectedwiththevertex?
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
in case of :(1/3) ,(S_(ABCD) /S_(MNPQ) )=2.5,(S_(ABCD) /S_(BNT) )=60.
incaseof:13,SABCDSMNPQ=2.5,SABCDSBNT=60.
Answered by ajfour last updated on 10/Jun/17
Commented by ajfour last updated on 10/Jun/17
all //gms that seem like the  central one are alike, which   is not so in diagram, but can  be judged so.   Then 5 such //gm areas=     Area of rectangle  The corner Δ′s have one-fourth  of the //gm area= one-twentieth  of the rectangle area.
all//gmsthatseemlikethecentralonearealike,whichisnotsoindiagram,butcanbejudgedso.Then5such//gmareas=AreaofrectangleThecornerΔshaveonefourthofthe//gmarea=onetwentiethoftherectanglearea.
Commented by ajfour last updated on 10/Jun/17
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
tg∠NBT=tg∠ADM=(b/(2a))⇒QM∥PN(1)  tg∠QCD=tg∠TAB=(a/(2b))⇒PQ∥MN(2)  from(1),(2)⇒MNPQ,is a parallelogram.
tgNBT=tgADM=b2aQMPN(1)tgQCD=tgTAB=a2bPQMN(2)from(1),(2)MNPQ,isaparallelogram.
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17

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