O is any point within a ∆ ABC whose sides are 4 cm, 5 cm, and 7 cm respectively. Prov that (OA+OB+OC) >8 cm.
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Answered by
19
The sum of any two sides of a triangle > the third side.
.
Hence OA + OB > AB
..........OB + OC > BC and
..........OC + OA > CA.
.
Adding all the three 2( OA + OB + O C ) > AB+BC+CA = 4+5+7 = 16
.
Hence OA+OB+OC > 8 .
HOPING IT WILL BE HELPFUL
.
Hence OA + OB > AB
..........OB + OC > BC and
..........OC + OA > CA.
.
Adding all the three 2( OA + OB + O C ) > AB+BC+CA = 4+5+7 = 16
.
Hence OA+OB+OC > 8 .
HOPING IT WILL BE HELPFUL
THANKS FOR MAKING ME BRAINLEST
Answered by
8
Answer:
Sum of two sides is greater than the third side.
OA+OB>4
OB+OC>5
OC+OA>7
2OA+ 2OB+ 2OC > 4+5+7
2 (OA+OB+OC) > 16
OA+OB+OC > 8
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