Question

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)

Answer 1

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)