Question

In Fig. the square ABCD is divided into five equal parts, all having same area. The central part is circular and the lines AE, GC, BF and HD lie along the diagonals AC and BD of the square. If AB = 22 cm, find:

Answer 1

GIVEN:
Side of a square (AB)= 22 cm
Lets the radius of the central part be r cm
Area of the central part= ⅕ × area of the square
πr²= ⅕(22×22)
22/7×r² =( 22×22)/5
r² = (22×7)/5= 154/5
r²= 30.8
r = √30.8 = 5.549≈ 5.55 cm
i) Circumference of the central part= 2πr
= 2×(22/7)×5.55= 34.88 cm

ii) Let O is the centre of the central part and O is also the centre of the square.
Diagonal of square AC= √2a
OA = OC = ½ AC (diagonals of a square are equal in length and bisect each other)
OA = √2a/2 = √2 ×22/2= 11√2
AE = BF= OA - OE = 11√2 - 5.55 = 11 × 1.41 - 5.55 = 15.51 - 5.55 = 9.96 cm
[√2= 1.41]
EF = ¼(circumference of the circle)= ¼(2πr)= πr/2 = 1/2×22/7× 5.55 = 8.72 cm
Perimeter of part ABEF = AB + AE +EF +BF= 22 + 9.96 + 8.72 + 9.96= 50.64 cm
Hence, the circumference of the central part=34.88 cm and the Perimeter of part ABEF =50.64 cm.