Question

ABCD is a trapezium with AB||DC, AB= 18 cm, DC= 32 cm and the distance between AB and DC is 14 cm. Circles of equal radii 7 cm with centres A, B, C and D have been drawn,Then, find the area of the shaded region of the figure.(π = 22/7).

Answer 1

From the GIVEN FIGURE, we have
AB = 18 cm , DC = 32 cm , Distance between AB  and DC(h) = 14 cm & radius of each circle(r) = 7cm

Since , AB ||DC
Therefore , ∠A + ∠D = 180° & ∠B + ∠C = 180°
[CO INTERIOR ANGLES are supplementary]

Area of sector =  (θ /360) ×πr²
Area of sector with ∠A & ∠D = (180 /360) × 22/7 × 7²
= ½ × 22 × 7 = 11 × 7 = 77 cm²

Similarly,  Area of sector with ∠B & ∠C = (180 /360) × 22/7 × 7²
= ½ × 22 × 7 = 11 × 7 = 77 cm²

Area of trapezium = ½ (sum of parallel sides) ×distance between Parallel sides(h)
Area of trapezium = ½ (AB + DC ) ×(h)
Area of trapezium = ½(18+32)×14
= ½(50)× 14 = 25 × 14 = 350 cm².

Area of shaded region = Area of trapezium - (Area of sector with ∠A & ∠D + Area of sector with ∠B & ∠C)
= 350 -(77+77) = 350 - 154 = 196 cm²
Area of shaded region = 196 cm²

Hence, the Area of shaded region is 196 cm².

HOPE THIS WILL HELP YOU...

Answer 2

Area of the shaded region in the fig.
                         =  area of the trapezium - area of the circles formed in fig.
                         = [(a+b)/2]h  - (πr²)            {here as the sum of all the angles of the trapezium is 360* therefore we can consider only one circle divided in 4 cornners respectively}
                           
                          = [(18+32)/2]×14 - 22/7× (7)²
                          = [50/2]×14  - 22×7
                          = (50 - 22)× 7
                          = 28 × 7 = 196 cm²