 ## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 29 of Exercise 10F

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29
##### The perimeter of a rectangle is 82 m and its area is 400 m2. The breadth of the rectangle is

Let the length and breadth of the rectangle be l and b respectively

Perimeter of a rectangle is 82 m

2(l + b) = 82

l + b = 41

l = 41 - b - - - - - - (1)

Area is 400 m2

lb = 400

(41 - b) b = 400 using (1)

41b –b2 = 400

b2 - 41b + 400 = 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 = 1 b = - 41 c = 400

= 1.400 = 400

And either of their sum or difference = b

= - 41

Thus the two terms are - 25 and - 16

Difference = - 25 - 16 = - 41

Product = - 25. - 16 = 400

b2 - 25b - 16b + 400 = 0

b(b - 25) - 16(b - 25) = 0

(b - 25) (b - 16) = 0

b = 25 or b = 16

If b = 25 l = 41 - 25 = 16 but l cannot be less than b

Thus b = 16m

The breadth of the rectangle = 16m

29

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