## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 65 of Exercise 10E

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65
##### A farmer prepares a rectangular vegetable garden of area 180 sq metres. With 39 metres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden.

Let the length and breadth of rectangular plot be x and y respectively.

Area = xy = 180 sq m – – – – – (1)

2(x + y) –x = 39

2x + 2y – x = 39

2y + x = 39

x = 39 – 2y

Putting the value of x in (1) we get

(39 – 2y)y = 180

39y – 2y2 = 180

2y2 – 39y + 180 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2 b = – 39 c = 180

= 2.180 = 360

And either of their sum or difference = b

= – 39

Thus the two terms are – 24 and – 15

Difference = – 24 – 15 = – 39

Product = – 24. – 15 = 360

2y2 – 24y – 15y + 180 = 0

2y(y – 12) – 15(y – 12) = 0

(y – 12)(2y – 15) = 0

y = 12 or y = 15/2 = 7.5

if y = 12 x = 39 – 2y = 39 – (2.12) = 39 – 24 = 15

if y = 7.5 x = 39 – 2y = 39 – [(2)(7.5)] = 39 – 15 = 24

Hence either l = 24 m, b = 7.5 m or l = 15 m, b = 12 m

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