**Answer:** Option **
B **

**Explanation:**

**Traditional Approach :**
Let the length of the playground be x and breadth be y.

Given, area of the rectangular playground = 100 square metre

∴ x × y = 100 ...(1)

30 m barbed wire is used to fence 3 sides. If one length side is not fenced using wire then we can say,

x + 2y = 30 ...(2)

⇒ x + 2 × |
100 |
= 30 [From eq. (1)] |

x |

⇒ x

^{2} + 200 = 30x

⇒ x

^{2} – 30x + 200 = 0

⇒ x

^{2} – 20x – 10x + 200 = 0

⇒ x(x – 20) – 10(x – 20) = 0

⇒ (x – 20)(x – 10) = 0

⇒ x = 20 or 10

Now, From eq. (1),

y = 5 or 10

As the playground is rectangular, x = y = 10 is not possible, so x = 20 m and y = 5 m.

____________________________________________________

**Intuitive Approach :**
The only two possible combinations of two sides of a rectangle, the are of which is 100 square metre are is 20 m × 5 m and 25 m × 4 m.

Assuming 20 m × 5 m to be true, we can validate our choice by putting the values in the following given condition.

x + 2y = 30 m

⇒ 20 + 2 × 5 = 30

⇒ 30 = 30

L.H.S. = R.H.S.

Hence, it is evident that the pair of sides (20 m × 5 m) is correct.