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16. A is any point within a trianglePQR whose sides are 5 cm, 7 cm, 10cm respectively. Prove thatPQ+QR+RP > 11​

Answer»

Answer:

Step-by-step explanation:

Referring to the figure attached below we have,

In ∆PQR,

PQ = 6 cm

QR = 7 cm

PR = 9 cm

Let O be any point INSIDE the triangle and join OP, OQ & OR.

According to the Triangle Inequality THEOREM, the sum of any TWO sides of a triangle must be greater than the measure of the third side.

Therefore,

In ∆POQ, [OP + OQ] > 6 cm ….. (i)

In ∆QOR, [OQ + OR] > 7 cm ….. (ii)

In ∆POR, [OP + OR] > 9 cm ……. (iii)

Now, adding eq. (i), (ii) & (iii), we get

[OP + OQ] + [OQ + OR] + [OP + OR] > [6 + 7 + 9] cm

⇒ 2 [OP + OQ + OR] > [22 cm]

⇒ [OP + OQ + OR] > [\frac{22}{2}

2

22

cm

⇒ [OP + OQ + OR] > 11 cm

Hence proved

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