NCERT Class 10 Mathematics

232 MATHEMATICS

Solution : Area of the square lawn ABCD = 56 × 56 m^2 (1)

Let OA = OB = x metres

So, x^2 + x^2 =56^2

or, 2 x^2 = 56 × 56

or, x^2 = 28 × 56 (2)

Now, area of sector OAB =^2

90

360

✁x =^2

1

4

✁x

=

(^122) 28 56 m 2
47
[From (2)] (3)
Also, area of ✂ OAB =^2

1

56 56 m

4

✄ ✄ (☎ AOB = 90°) (4)

So, area of flower bed AB =

(^122) 28 56 (^1) 56 56 m 2
47 4

✆ ✞ ✞ ✞ ✟ ✞ ✞ ✝

✠ ✡

☛ ☞

[From (3) and (4)]

=

(^12) 28 56 (^2) 2 m 2
47

✎ ✎ ✌ ✏ ✍

✑ ✒

✓ ✔

=^2

18

28 56 m

47

(5)

Similarly, area of the other flower bed

=

(^18) 28 56 m 2
47

✄ ✄ ✄ (6)

Therefore, total area =

18

56 56 28 56

47

✆ ✞ ✕ ✞ ✞ ✞

✠

☛

(^182856) m 2
47

✕ ✞ ✞ ✞ ✝

✡

☞

[From (1), (5) and (6)]

=

28 56 2^22 m^2

77

✎ ✌ ✖ ✖ ✍

✑ ✒

✓ ✔

=^22

18

28 56 m 4032 m

7

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