Question

Question 17 Answer correctly​

Answer 1

Answer:

Given,

Side of the square = 22 cm

= AB Let the radius of the centre part be r cm.

Then, area of the circle = 1/5 x area of the square πr2

= 1/5 x 222 22/7 x r2

= (22 x 22)/ 5 r

= 154/5 = 5.55 cm

(i) Circumference of central part = 2πr = 2(22/7)(5.55)

= 34.88 cm.

(ii) Let O be the center of the central part.

Then, its clear that O is also the center of the square as well.

Now, in triangle ABC By Pythagoras theorem AC2 = AB2 + BC2 = 222 + 222 = 2 x 222 AC = 22√2

Since diagonals of a square bisect each other AO = 1/2AC = 1/2 (22√2) = 11√2 cm

And, AE = BF = OA – OE = 11√2 – 5.55 = 15.51 – 5.55 = 9.96 cm

EF = 1/4(Circumference of the circle) = 2πr/4 = 1/2 x 22/7 x 5.55 = 8.72 cm

Thus, the perimeter of the part ABEF = AB + AE + EF + BF

= 22 + 2 x 9.96 + 8.72

= 50.64 cm