• Word Problem based on Geometric figure (Rectangle)

QUADRATIC �EQUATIONS

–5x²

+ 2940

– 35x

– 5x

– 21x

– 588

= 0

(x + 28)

– 21

= 0

+ 28x

+ 7x

– 588

= 0

(x + 28)

1

=

420

x

+

7

A rectangular playground is 420 sq.m. If its length is increased by 7m and breadth is decreased by 5 metres, the area remains the same.

Find the length and breadth of the playground ?

12.

Sol.

Let the length of a rectangular playground be ‘x’ m.

Area of rectangle = length × breadth

420 =

= breadth

×

=

420

=

420

420

+

– 35

Multiplying throughout by x, we get;

0 =

5x² + 35x – 2940 = 0

Dividing throughout by 5

x²

x²

x

(x + 28)

(x – 21)

= 0

x + 28 = 0

or x – 21 = 0

x = – 28

or x = 21

The length of playground cannot be negative

x ≠ – 28

Hence, x = 21

Breadth

∴ The length of rectangular playground is 21m and its breadth is 20m.

What do we have to find ?

We know that

420 sq.m

x m

x

+

7

m

(

)

We are considering a rectangular playground

Length and Breadth are unknown

Area of new rectangle = length × breadth

Let’s do the prime factorisation of 588

588

294

147

2

2

3

49

7

7

7

1

+

–

14

–

42

21

–

= 7

28

x

× breadth

Since we are subtracting the factors give middle term sign to the bigger factor and the opposite sign to the smaller factor

Find two factors of 588 in such a way that by subtracting factors we get middle no. 7