In fig(iv), T is a point on side QR of triangle PQR and S is a point such that RT = ST. Prove that PQ + PR > QS

Open in App

Solution

In $△ P Q R, P Q + P R > Q R$ $∵$ sum of the two sides is greater than the third side.
$∴ P Q + P R > Q T + T R$ $∵ Q R = Q T + T R$
$\Rightarrow P Q + P R > Q T + T S$ ..... $(1)$ $∵ T R = T S$
In $△ Q S T$ , $Q T + T S > Q S$ ..... $(2)$
$∴$ from $(1)$ and $(2)$ we have
$P Q + P R > Q S$
Hence proved

Suggest Corrections

Similar questions

In Fig. T is a point on side QR of △PQR and S is a point such that RT = ST. Prove that PQ + PR > QS

In Fig. T is a point on side QR of $Δ P Q R$ and S is a point such that RT = ST. Which of the following is true?

In the given figure, T is a point on side QR of $Δ P Q R$ and S is a point such that RT = ST.
Which of the following is true?

S

T

P R

Q R

△ P Q R

∠ P = ∠ R T S

△ R P Q \sim △ R T S