# ABCD is a quadrilateral. Is AB+BC+CD+DA < 2(AC+BD

OD+OC>DC

OB+OC>BC

AO + OB + OA + OD + OC + OB + OC >  AB + AC + BC + CD

2(AC+BD)>AB+AC+BC+CD

HENCE YES THE STATEMENT IS TRUE

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yes

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AO+OB AB

OD+OC DC

OD+OC DC

OB+OC BC

AO+OB+OA+OD+OC+OB+OD+OC AB+AC+BC+CD

proved

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2AC AB + BC Similar way, 2AC CD + DAAlso, 2BD DA + AB and 2BD BC + CDAdding the inequalities, we get 4(AC + BD) 2(AB + BC +CD +DA)Dividing by 2 on both sides,we arrive at 2(AC + BD) (AB + BC +CD +DA)or (AB + BC +CD +DA)

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AB+BC+CD+DA2AC

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AO + OB AB
OD + OC DC
OB + OC BC

AO+OB+OA+OD+OC+OB+OB+OD+OC AB+BC+CD

YES,THIS STATEMENT IS TRUE
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IN THE EARLIER QUESTION WHERE 2 WAS NOT THR IN AC WE DIDNT TOUCH THAT O NOW WHEN 2 AC IS THERE WE ARE USING O CN U SXPLAIN THAT

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DONT KNOW

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I cannot understand you.
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Repaet the question
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You will get the answer of this question in the column of NCERT Solutions.
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YES IT IS TRUE
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yes
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njkk kh.

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STATEMENT IS TRUE DEAR..
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Since, the sum of lengths of any two sides in a triangle should be greater than the length of the third side.

Therefore,

In Δ PQR, PQ + QR > PR – – – – – – – – – – (1)

In Δ PSR, PS + SR > PR – – – – – – – – – (2)

In Δ SRQ, SR + RQ > SQ – – – – – – – – – (3)

In Δ PSQ, PS + PQ > SQ – – – – – – – – – – (4)

Adding the equations (1) , (2) , (3) and (4) we get,

PQ + QR + PS + SR + RQ + PS + PQ > PS + PS + SQ + SQ

=>( PQ + PQ ) + ( QR + QR ) + ( PS + PS ) + ( SR + SR ) > 2 PR + 2 SQ

=> 2 PQ + 2 QR + 2 PS + 2 SR > 2 PR + 2 SQ

=> 2 ( PQ + QR + PS + SR) > 2 ( PR + SQ )

=> PQ + QR + PS + SR > PR + SQ

=> PQ + QR + RS + SP > PR + SQ

Hence, it is true.

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Draw a triangle PQR in which PQ =3cm QR =4cm RP=5cm . Also draw prependicular bisector of side PQ.
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SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER TAN THE THIRD SIDE
IN TRIANGLE ABC
AB+BC IS GREATER THAN AC
CD+DA IS GREATER THAN AC
THUS AB+BC+CD+DA IS GREATER THAN 2AC
IN TRIANGLE BDC
CD+BC IS GREATER THAN BD
THUS AB+BC+CD+DA IS GREATER THAN 2BD
THUS AB+BC+CD+DA+AB+BC+CD+DA IS GREATER THAN 2(AC + BD)
= 2(AB+BC+CD+DA) IS GREATER THAN 2(AC +BD)
THUS AB+BC+CD+DA IS GREATER THAN AB+BD
HENCE IT PROVED
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yes, the statement is indeed correct

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yes
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yes its true

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• SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER TAN THE THIRD SIDE

IN TRIANGLE ABC

AB+BC IS GREATER THAN AC

CD+DA IS GREATER THAN AC

THUS AB+BC+CD+DA IS GREATER THAN 2AC

IN TRIANGLE BDC

CD+BC IS GREATER THAN BD

THUS AB+BC+CD+DA IS GREATER THAN 2BD

THUS AB+BC+CD+DA+AB+BC+CD+DA IS GREATER THAN 2(AC + BD)

= 2(AB+BC+CD+DA) IS GREATER THAN 2(AC +BD)

THUS AB+BC+CD+DA IS GREATER THAN AB+BD

HENCE IT PROVED
•
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NO IT CANNOT
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50 is the interior angle
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Angry teri maa ko chodu
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Hence it is proved see draw a quadilater mark a point o n separate it into 4 sides
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I AGGREA DEVA

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Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side Therefore,  In Δ AOB, AB < OA + OB ……….(i)  In Δ BOC, BC < OB + OC ……….(ii)  In Δ COD, CD < OC + OD ……….(iii)  In Δ AOD, DA < OD + OA ……….(iv)  ⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD  ⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]  ⇒ AB + BC + CD + DA < 2(AC + BD)  Hence, it is proved.

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OD+OC>DC

OB+OC>BC

AO + OB + OA + OD + OC + OB + OC > ?AB + AC + BC + CD

2(AC+BD)>AB+AC+BC+CD

HENCE YES THE STATEMENT IS TRUE
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A figure in which four right angles & equal to each other is known as square
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And in some cases equal
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Ab+bc+cd+de≠2*(ab+cd
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OD+OC>DC

OB+OC>BC

AO + OB + OA + OD + OC + OB + OC >  AB + AC + BC + CD

2(AC+BD)>AB+AC+BC+CD

HENCE YES THE STATEMENT IS TRUE

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decimal multiply fraction
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The Statement is true.
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