Question

From a thin metallic piece, in the shape of a trapezium ABCD, in which AB is parallel to CD and ∠BCD=90∘, a quarter circle BEFC is removed. Given AB=BC=3.5cm and DE=2cm, calculate the area of the remaining piece of the metal sheet.

Answer 1

Answer:

The area of remaining piece of the metal sheet (Shaded region) is 6.125 cm².

Step-by-step explanation:

Given :

In trapezium ABCD in which AB || CD and ∠BCD = 90°  

AB = BC =3.5 cm & DE = 2 cm  

 

CE = CB = 3.5 cm  

[CE and BC are the radii of quarter circle BFEC]

so,DC = DE + EC  

DC = 2 cm + 3.5 cm  

DC = 5.5 cm  

Area of remaining piece of the metal sheet (Shaded region), A = Area of trapezium ABCD - Area of quarter circle BFEC  

= ½ (AB + DC) × BC - ¼ π(BC)²

[Area of trapezium = ½ (sum of parallel sides) × perpendicular distance between the parallel sides & area of quadrant = ¼ π r²]

A = ½ (3.5 + 5.5) × 3.5  - ¼ π(3.5)²

A = ½ × 9  × 3.5 - ¼ π(3.5)²

A = 4.5 × 3.5  - 22/7 × 3.5 × 3.5/4

A = 15.75 - ½ × 11 × 0.5 × 3.5

A = 15.75 - 19.25/2

A = 15.75 - 9.625

A = 6.125 cm²  

Area of remaining piece of the metal sheet (Shaded region) = 6.125 cm²

Hence, the area of remaining piece of the metal sheet (Shaded region) is 6.125 cm².

Answer 2

Refer The Attachment ⬆️

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