Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 29 of Exercise 10F
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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
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