In any quadrilateral, the sum of the lengths of four sides is greater than the two diagonals. Prove.
Answers
Answered by
4
Answer:
Let PQRS be the given quadrilateral.
In ∆PQR,
PQ + QR > PR ...(1)(Sum of any to sides of a triangle is greater than third side)
In ∆RSP,
RS + SP > PR ...(2)
In ∆PQS,
PQ + SP > QS ...(3)
In ∆QRS,
QR + RS > QS ...(4)
By adding (1), (2), (3) and (4),
2(PQ + QR + RS + SP) > 2(PR + QS)
=> PQ + QR + RS + SP > PR + QS
Hence, proved.
Thanks!!
Please mark brainliest and follow me...
Attachments:
Similar questions