In Fig. PQRS is a quadrilateral in which diagonals PR and QS intersect in O. [4 MARKS]
Show that:
(i) PQ + QR + RS + SP >PR + QS
(ii) PQ + QR + RS + SP <2 (PR + QS)
Sub-parts : 2 Marks each
Since the sum of any two sides of a triangle is greater than the third side.
Therefore, in △PQR, we have
PQ+QR > PR ......(i)
In △ RSP, we have
RS +SP > PR ....(ii)
In △ PQS, we have
PQ + SP > QS .....(iii)
In △ QRS, we have
QR + RS > QS ......(iv)
Adding (i), (ii), (iii) and (iv), we get
PQ + QR + RS + SP > PR + QS
This proves (i)
Now in △OPQ, we have
OQ + OP > PQ ...(v)
In △OQR, we have
OQ + QR > QR ...(vi)
In △ORS, we have
OR + OS > RS ....(vii)
In △OSP, we have
OS + OP > SP ....(viii)
Adding (v), (vi), (vii) and (viii), we get
PQ + QR + RS + SP < 2 (PR + QS)