Question

Construct a cyclic quadrilateral EFGH with EF = 6 cm,$\angle GEF={60}^{\circ }$, FG = 7cm and $\angle EGH={50}^{\circ }$.

Answer 1

In the Cyclic Quadrilateral EFGH

EF = 6 cm, $\angle GEF={60}^{\circ }$, FG = 7cm and $\angle EGH={50}^{\circ }$.

Construction:

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

(ii) From E, draw EX such that $\angle FEX={60}^{\circ }$.

(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 $\angle EGY={50}^{\circ }$ which intersects the circle at H.

(viii) Join EH.

Now, EFGH is the required cyclic quadrilateral.