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From a Thin Metallic Piece, in the Shape of a Trapezium Abcd, in Which Ab || Cd and ∠Bcd = 90°, a Quarter Circle Befc is Removed (In the Following Figure). Given Ab = Bc = 3.5 Cm - Mathematics


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

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We have given a trapezium. We are asked to find the area of the shaded region.

We can find the area of the remaining part that is area of the shaded region as shown below.

`"Area of the shaded region = Area of the trapezium-Area of the sctor"` 

`∴ "Area of the shaded region"=1/2 (AB+CD)xxBC-θ/360 pir^2`

`∴ "Area of the shaded region"=1/2(3.5+CD)xx3.5-90/360 pi(3.5)^2 .............(1)` 

Now we find the value of CD. 


`∴ CD=3.5+2` ..................(Since, CE is radius of the sector, therefore, CE = 3.5)

`∴ CD=5.5`

Substituting the values of CD and `pi=22/7`in equation (1), 

`∴ "Area of the shaded region" = 1/2(3.5+5.5)xx3.5-90/360xx22/7xx(3.5)^2`

`∴ "Area of the shaded region"=31.5/2-1/4xx22xx0.5xx3.5`

`∴ "Area of the shaded region"=31.5/2-1/2xx11xx0.5xx3.5`

`∴ "Area of the shaded region"=31.5/2-19.25/2`

`∴ "Area of the shaded region"=12.25/2`

`∴ "Area of the shaded region"=6.125`

Therefore, area of the remaining part is `6.125 cm^2`

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RD Sharma Class 10 Maths
Chapter 13 Areas Related to Circles
Exercise 13.4 | Q 49 | Page 64
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