Question

ABCD is a trapezium with ab parallel to dc and angle abc is equal to 90 degree BC is a quadrilateral of a circle if BC = CD = 4.2 cm and ae = 1.8 cm find the area of shaded region​

Answer 1

Answer:

ANSWER

Given: CE=CB=7cm

∴ CD=CE+ED=(7+4)cm=11cm

In Δ CLB, we have

sin60

∘

=

BC

BL

⇒

2

3

=

7

BL

⇒BL=

2

7

3

cm

∴ Area of trapezium =

2

1

(AB+CD)×BL=

2

1

(7+11)×

2

7

3

cm

2

=

2

63

3

cm

2

Area of sector BFEC =

360

∘

60

∘

×

7

22

×7

2

cm

2

=

3

77

cm

2

.......(Area of sector=

360

θ

×πr

2

)

∴ Required area = (

2

63

3

−

3

77

)cm

2

=(54.558−25.666)cm

2

=28.89cm

2

Answer 2

Step-by-step explanation:

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Given: CE=CB=7cm

∴ CD=CE+ED=(7+4)cm=11cm

In Δ CLB, we have

sin60

∘

=

BC

BL

⇒

2

3

=

7

BL

⇒BL=

2

7

3

cm

∴ Area of trapezium =

2

1

(AB+CD)×BL=

2

1

(7+11)×

2

7

3

cm

2

=

2

63

3

cm

2

Area of sector BFEC =

360

∘

60

∘

×

7

22

×7

2

cm

2

=

3

77

cm

2

.......(Area of sector=

360

θ

×πr

2

)

∴ Required area = (

2

63

3

−

3

77

)cm

2

=(54.558−25.666)cm

2

=28.89cm

2