Question

in a triangle ABC,AB=5cm,BC=6cm,CA=7cm.if 'o' is any point inside the triangle then oA+oB+oc is greater than?

Answer 1

Answer:

a triangle ABC, AB=5cm,BC=6cm,CA=7cm . If O is any point inside the triangle, then OA+OB+OC is greater than

Answer 2

• in ∆ABC, when AB = 5cm, BC = 6cm, CA = 7cm and 'o' is any point inside the triangle then (OA + OB + OC) is greater than 9 cm .

Given :- in a triangle ABC, AB = 5cm, BC = 6cm, CA = 7cm. 'o' is any point inside the triangle .

To Find :-

• (OA + OB + OC) is greater than ?

Concept used :-

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

Solution :-

From image we can see that, ABC is a triangle and O is a point inside it. Join AO, BO and CO .

Now, In ∆AOB we have,

→ AO + BO > AB { since sum of any two sides is greater than the third side . } ------------ Equation (1)

Similarly, In ∆BOC we have,

→ BO + CO > BC ------------ Equation (2)

also, In ∆AOC we have,

→ AO + CO > AC ------------ Equation (3)

adding all three equations we get,

→ (AO + BO) + (BO + CO) + (AO + CO) > AB + BC + AC

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

→ 2AO + 2BO + 2CO > AB + BC + AC

→ 2(AO + BO + CO) > AB + BC + AC

dividing both sides by 2,

→ (AO + BO + CO) > (AB + BC + AC)/2

we can write this as,

→ (OA + OB + OC) > (AB + BC + AC)/2

putting given values of AB, BC and AC now,

→ (OA + OB + OC) > (5 + 6 + 7)/2

→ (OA + OB + OC) > (18/2)

→ (OA + OB + OC) > 9 cm (Ans.)

Therefore, we can conclude that, (OA + OB + OC) is always greater than 9 cm .

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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