# 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

Sum

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.

#### Solution

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=CE+DE

∴ 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=22/7in 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

Is there an error in this question or solution?

#### APPEARS IN

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