Question

ABCD is a rectangle of length 20 cm and breadth 10√2cm. OAPB is a sector
of a circle of radius 102 cm. Calculate the area of the shaded region.
[ Take = 3:14 ]
​

Answer 1

Answer:

143 cm^2

Step-by-step explanation:

Construct/complete the square(of side length 20 cm).

Notice that O is on the line of perpendicular bisector of DC,  which means O is the mid point of A'B'    ∴ OA' = OB' = 10 cm

Moreover, as DCB'A' is a square, side length = 20 cm

      ∴ AD + AA' = 20     &  BC + BB' = 20

       ∴ AA' = 10       & BB = 10

    Now,

As in Δ AA'O:    ∵ AA' = A'O = 10 cm

                          ∴ ∠A'AO = ∠AOA' = x(say)

And, ∠AA'O = 90°

    ∴ 90° + x + x = 180°       ⇒ x = 45°

    ∴ ∠AOA' = 45°

Similarly in triangle BB'O:   ∠BOB' = 45°

     

  Hence, ∠AOB = 180° - 45° - 45°

                                  = 90°                    

which clearly means, AOBP forms a quarter circle and area should be 1/4 of πr².            Hence, area of sector = 1/4 * 3.14 * (10√2)² = 157 cm².

Observing,

Shaded region = area of square - area of sector - area of ΔAA'O - area of ΔBB'O

         = (20)² - 157 - (1/2)(10)(10) - (1/2)(10)(10)

         = 400 - 157 - 50 - 50

         = 143 cm^2

[units have been ignored (wherever possible) ;  area of triangle = 1/2 * base * height     ; area of square = side^2    ;   area of full circle = πr^2]