CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In the given figure, PQR is a triangle and S is any point in its interior. Which of the following options is correct?

  1. SQ + SR > PQ + PR
  2. SQ + PQ > SR + PR
  3. SQ + SR < PQ + PR
  4. SQ - SR > PQ - PR


Solution

The correct option is C SQ + SR < PQ + PR
S is any point in the interior of Δ PQR.
In Δ PQT,PQ+PT>QT[ Sum of the two sides of a triangle is greater than the third side]
PQ+PT>SQ+ST
[QT=SQ+ST]...(1)
Also, in Δ RST,
ST + TR > SR...(2)[ Sum of the two sides of a triangle is greater than the third side]
Adding (1) and (2), we get 
PQ + PT + ST + TR > SQ + ST + SR
PQ+(PT+TR)>SQ+SR
[Cancelling ST on both sides.]
PQ+PR>SQ+SR 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More



footer-image