Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

pratu043

  • 3 years ago

If S is any point in the interior of triangle PQR then prove that SQ + SR < PQ + QR.

  • This Question is Closed
  1. pratu043
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1345956288365:dw|

  2. pratu043
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    someone please help!!

  3. pratu043
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    please!!!

  4. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry can u check it plz... i have some doubt on it SQ + SR < PQ + QR or SQ + SR < PQ + PR

  5. pratu043
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry its SQ + SR < PQ + PR.

  6. jiteshmeghwal9
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Does it help @pratu043 ??? \[\LARGE{s=90^o+{1 \over 2} \angle a}\]

  7. SNSDYoona
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    was wondering if i could use vectors to proof it? or u want basic trigonometries?

  8. TheMind
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the perimeter of tri PQR > the perimeter of tri SQR because S is inside PQR. in other word PQ+QR+PR > SQ+QR+SR take out the positive amount QR from both side PQ+PR > SQ+SR

  9. pratu043
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you @TheMind

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy