construct trapezium EFGH in which EF is parallel to GH, EF=5 cm, FG 7 cm and angle f = 60 - Maths - Practical Geometry

Question

construct trapezium EFGH in which EF is parallel to GH, EF=5 cm, FG
7 cm and angle f = 60 GH=6cm

Vijay Kumar Gupta · Accepted Answer

Use the following steps of construction to draw the required trapezium
1) Draw a line EF with measure 5 cm
2) At F, draw an angle, ∠XFE equal to 60°3) From the arc FX, cut of a line segment FG=7 cm4) Since the sum of interior angles on the same side of transversal equals 180°
To draw a line parallel to EF, at point G, draw an angle, ∠YGF =120°From this ray GY, cut of a line segment GH=6 cm
Join E and H to get the required trapezium
The figure is shown below:

construct trapezium EFGH in which EF is parallel to GH, EF=5 cm, FG 7 cm and angle f = 60 GH=6cm