One of the important question for the board exam class 10 from

mensuration chapter.This question is from r d sharma class 10 maths.

chapter-15.4 question number-17 maths r d sharma,mensuration.

In this question we have to find the circumference of the central

part and the perimeter of the part ABEF given that 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.

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 = ½ AC = ½ (22√2) = 11√2 cm

And,

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

EF = ¼(Circumference of the circle) = 2πr/4

= ½ 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