Question

Construct a cyclic quadrilateral EFGH with EF$=5.2$cm, $\angle$GEF$={50}^o$, FG$=6$cm and $\angle$EGH$={40}^o$.

Answer 1

Given: In the Cyclic Quadrilateral EFGH
EF$=5.2$cm, $\angle$GEF$={50}^o$, FG$=6$cm and $\angle$EGH$={40}^o$.
Construction:
(i) Draw a rough diagram and mark the measurements. Draw a line segment EF$=5.2$cm.
(ii) From E, draw EX such that $\angle$FEX$={50}^o$.
(iii) With F as centre and radius $6$cm, draw an arc intersecting EX at G.
(iv) Join FG.
(v) Draw the perpendicular bisectors of EF and FG intersecting each other at O.
(vi) With O as centre and OE(=OF=OG) as radius, draw a circumcircle.
(vii) From G, draw GY such that $\angle$EGY$={40}^o$ which interscets the circle at H.
(viii) Join EH.
Now, EFGH is the required cyclic quadrilateral.