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

IN A QUADILATERAL AO+OB>AB(INEQUALITY)

OD+OA>AD

OD+OC>DC

OB+OC>BC

ADDING

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

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

HENCE YES THE STATEMENT IS TRUE

  • 56

yes

  • -8

AO+OB AB

OD+OC DC

OD+OC DC

OB+OC BC

now Adding these

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

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

proved

  • -5

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)

  • 2

AB+BC+CD+DA2AC

  • -10
AO + OB AB
OD + OC DC
OB + OC BC

NOW ADDING THESE 
AO+OB+OA+OD+OC+OB+OB+OD+OC AB+BC+CD
2(AC+BD) AB+BC+CD+AD

YES,THIS STATEMENT IS TRUE
  • -1
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
 
  • 5
DONT KNOW
 
  • -6
I cannot understand you.
  • -3
Repaet the question
  • -9
You will get the answer of this question in the column of NCERT Solutions.
  • -4
YES IT IS TRUE
  • -8
yes
  • -6
njkk kh.

 
  • -6
STATEMENT IS TRUE DEAR..
  • 10

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

13

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.

  • 3
Draw a triangle PQR in which PQ =3cm QR =4cm RP=5cm . Also draw prependicular bisector of side PQ.
  • -3
Please find this answer

  • -2
SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER TAN THE THIRD SIDE
IN TRIANGLE ABC
AB+BC IS GREATER THAN AC
IN TRIANGLE ADC
CD+DA IS GREATER THAN AC
THUS AB+BC+CD+DA IS GREATER THAN 2AC
IN TRIANGLE ADB
AD+AB IS GREATER THAN BD
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
  • 0
yes, the statement is indeed correct
 
  • 0
yes 
  • 1
yes its true
 
  • -1
Please find this answer

  • -2
Please find this answer

  • 0
Please find this answer

  • 2
what is capital of Australia
  • 0
Please find this answer

  • 0
Please find this answer

  • -1
  • SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER TAN THE THIRD SIDE

    IN TRIANGLE ABC

    AB+BC IS GREATER THAN AC

    IN TRIANGLE ADC

    CD+DA IS GREATER THAN AC

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

    IN TRIANGLE ADB

    AD+AB IS GREATER THAN BD

    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
  •  
  • 1
Help me to do

  • -1
Please find this answer

  • 0
NO IT CANNOT
  • 0
50 is the interior angle
  • 0
Angry teri maa ko chodu
  • 0
Hence it is proved see draw a quadilater mark a point o n separate it into 4 sides
  • 0
I AGGREA DEVA
 
  • 0
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.
 
  • 1
IN A QUADILATERAL AO+OB>AB(INEQUALITY)

OD+OA>AD

OD+OC>DC

OB+OC>BC

ADDING

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

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

HENCE YES THE STATEMENT IS TRUE
  • 0
A figure in which four right angles & equal to each other is known as square
  • 0
In some cases yes
In some cases no
And in some cases equal
  • 0
Ab+bc+cd+de≠2*(ab+cd
  • 0
11+1
  • 0
nice
  • 1
addition doubt
  • 0
correct
  • 0
Please find this answer

  • 0
HEY MATE THIS IS YOUR ANSWER
 
  • 0

IN A QUADILATERAL AO+OB>AB(INEQUALITY)

OD+OA>AD

OD+OC>DC

OB+OC>BC

ADDING

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

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

HENCE YES THE STATEMENT IS TRUE

  • 0
decimal multiply fraction
  • 0
The Statement is true.
  • 1
Dear student here is your answer:

  • 0
I give your answer

  • 0
I give your answer.

  • 1
Pubj
  • 0
Please find this answer

  • 0
Yes correct
  • 0
answers are correct 
  • 0
it is true
  • 0
yes to yes
  • 0
Please find this answer

  • 0
15 cm
  • 0
Video chapter animal kingdom biology explained with video
  • 0
IN A QUADILATERAL AO+OB>AB(INEQUALITY)

OD+OA>AD

OD+OC>DC

OB+OC>BC

ADDING

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

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

HENCE YES THE STATEMENT IS TRUE
  • 0
Ac + bd = acbd
  • 0
Please find this answer

  • 0
73503 48665
  • 0
Please find this answer

  • 0
What are you looking for?