Question

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.

Answer 1

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

Answer 2

OA + OB + OC > 8 cm.

Step-by-step explanation:

We are given that O is any point within an ∆ABC whose sides are 4 cm, 5 cm, and 7 cm respectively.

Let the side AB of the triangle ABC = 4 cm

the side BC of the triangle ABC = 7 cm

the side AC of the triangle ABC = 5 cm  

As we know that O is any point inside the ∆ABC and also it is stated that the sum of two sides of the triangle must be greater than the third side of the triangle.

SO, in the ∆OAB;

OA + OB > AB  

OA + OB > 4 cm   ------------- [equation 1]

In the ∆OBC;

OB + OC > BC  

OB + OC > 7 cm   ------------- [equation 2]

In the ∆OAB;

OA + OC > AC  

OA + OC > 5 cm   ------------- [equation 3]

Now, adding all the three equations we get;

(OA + OB) + (OA + OB) + (OA + OB) > 4 cm + 7 cm + 5 cm

2OA + 2OB + 2OC > 16 cm

2(OA + OB + OC) > 16 cm

OA + OB + OC > $\frac{16}{2}$

So,  OA + OB + OC > 8 cm

Hence proved.