Question

Construct a cyclic quadrilateral PQRS with PQ = 6 cm, $\angle P={105}^{\circ }$ , $\angle PQS={45}^{\circ }$ and $\angle SQR={75}^{\circ }$.

Answer 1

Given in the cyclic quadrilateral PQRS,

PQ = 6 cm, $\angle P={105}^{\circ }$ , $\angle PQS={45}^{\circ }$ and $\angle SQR={75}^{\circ }$.

Construction:

(i) Draw a line segment PQ = 6 cm.

(ii) From P draw PX such that $\angle QPX={105}^{\circ }$.

(iii) From Q draw QY such that $\angle PQY={45}^{\circ }$. Let QY meet PX at S.

(iv) Draw perpendicular bisectors of PQ and PS intersecting each other at O.

(v) With O as centre and OP( = OQ = OS ) as radius, draw a circumcircle of $\Delta PQS$

(vi) From Q, draw QZ such that $\angle SQZ={75}^{\circ }$ which intersects the circle at R.

(vii) Join RS.

Now, PQRS is the required cyclic quadrilateral.