5.1 INTRODUCTION
Sports committee of Dhannur High School wants to construct a Kho-Kho court of
dimensions 29 m × 16 m. This is to be
laid in a rectangular plot of area 558 m2.
They want to leave space of equal width
all around the court for the spectators.
What would be the width of the space
for spectators? Would it be enough?
Suppose the width of the space be
x meter. So, from the figure the length of
the plot would be (29 + 2x) meter.
And, breadth of the rectangular plot would bep = (16 + 2x) m.
Therefore, area of the rectangular plot = length × breadth
= (29 + 2x) (16 + 2x)
Since the area of the plot is
= 558 m2
(29 + 2x) (16 + 2x)
= 558
4x2 + 90x + 464
= 558
x2 + 90x - 94 =
=0
(dividing by 2)
2x2 + 45x - 47
= 0
2x2 + 45 x - 47
= 0
In previous class we solved the linear equations of the form ax + b = c to find the value of
‘x’. Similarly, the value of x from the above equation will give the possible width of the space for
spectators.
Can you think of more such examples where we have to find the quantities, like in the
above example and get such equations.
Let us consider another example:
Rani has a square metal sheet. She removed squares of side 9 cm from each corner of this
sheet. Of the remaining sheet, she turned up the sides to form an open box as shown. The
capacity of the box is 144 cm3
. Can we find out the dimensions of the metal sheet?
5 Quadratic Equations
