Question

3. In the adjoining figure, ABCD is a
rectangular piece of paper. It is cut along
the line CE that makes an angle of 30°
with the side BC. The sides of the piece of
paper are CD = 80 cm and BC = 60 cm.
Find the area of the bigger part between
the two parts.​

Answer 1

Answer:

area of bigger par - area of square - area of triangle. .

Step-by-step explanation:

area of square - l×b

-80×80=. 1600cm²

area of triangle - sin60/2 ×b×h

-. √3/2 ×60×80.

- 240√3

Answer 2

Step-by-step explanation:

In triangle CEB,

tan 30=BE/CB

1/under root 3=BE/60

60/under root 3=BE

BE=20×under root 3

BE=34.4cm

Area of the triangle CEB,

=1/2×BE×CD

=1/2×34.4×60

=34.4×30

=1032cm^2

Area of the rectangle,

=L×B

=80×60

=4800cm^2

Area of the bigger part,

=area of the rectangle-area of the triangle

=(4800-1032)cm^2

=3768cm^2