Question

Construct a cyclic quadrilateral PQRS with PQ$=4$cm, $\angle$P$={100}^o$, $\angle$PQS$={40}^o$ and $\angle$SQR$={70}^o$.

Answer 1

In the cyclic quadrilateral PQRS,
PQ$=4$cm, $\angle$P$={100}^o$, $\angle$PQS$={40}^o$ and $\angle$SQR$={70}^o$.
Construction:
(i) Draw a rough diagram and mark the measurements.
(ii) From P draw PX such that $\angle$QPX$={100}^o$.
(iii) From Q draw QY such that $\angle$PQY$={40}^o$. 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$={70}^o$ which intersects the circle at R.
(vii) Join RS.
Now, PQRS is the required cyclic quadrilateral.