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
Since, the sum of lengths of any two sides in a triangle should be greater than the length of the third side.
Therefore,
In
In
In
In
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
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
- 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
- 1