Math, asked by subhamcom999, 5 months ago

22) In the given figure, O is a point in the exterior of
a triangle ABC. O is joined to the three vertices
A, B and C. Show that OA + OB + OC >1\2(AB+BC+CA)


Answers

Answered by SweetCandy10
28

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Given -

O is a point in the exterior of ABC.

To prove -

OA + OB + OC > ½ (AB + BC + CA)

Construction -

Join OA, OB and OC

Property Used -

Sum of two sides in a triangle is greater than the third side .

Proof -

Using the above mentioned property - .

AO + OC > AC

AO + OB > AB

OB + OC > BC

Adding these three Inequalities -

2AO + 2BO + 2CO > AB + BC + AC

2 ( OA + OB + OC ) > ( AB + BC + AC )

( OA + OB + OC ) > ½ ( AB + BC + AC )

Hence proved

___________

Additional information -

  • 1. Sum of two sides of a triangle is always greater than the third side .

  • 2. Difference of two sides of a triangle is always less than the third side .

__________________

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Answered by Anonymous
6

Answer:

Given data:

O is a point in the exterior of triangle ABC

To prove: 2 [OA + OB + OC] > AB + BC + CA

We know that the sum of any two sides of a triangle is greater than the third side. Therefore, from the figure attached below, let us consider:

1. In ∆ OAB,

(ii) OA + OB > AB ....... (i)

2. In ∆ OCB,

(i) OB + OC > BC....... (ii)

3. In ∆ OAC,

OA + OC > CA ........ (iii)

Now, adding equation (i), (ii) & (iii), we get

[OA + OB + OB + OC + OA + OC] >

[AB + BC + CA]

Or, [(2*OA) + (2*OB) + (2*OC)] > [AB + BC + CA]

Or, 2 [OA + OB + OC] > [AB + BC + CA]

Hence proved

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