Question: The area of a rectangular field is 720 sq.m. and perimeter is 108 m. By what percent is the longer side of the field to be decreased to make it a square? Why?

Solution:
Given,

Area of a rectangular field (A) = 720 m²
Perimeter of the rectangular field (P) = 108 m

Let 'l' represent the length of the field and 'b' represent the breadth.

We have,
P = 108 m
$or, 2(l+b) = 108$
$or, 2(l+b) = 2×54$
$or, l +b = 54$
$or, l = 54 -b$ - (i)

Also,
A = 720 m²
$or, lb = 720$
[ Put value of l from equation (i) ]
$or, (54 - b)b = 720$
$or, 54b -b² = 720$
$or, b² -54b +720 = 0$
$or, b² -(30+24)b +720 = 0$
$or, b² - 30b -24b +720 = 0$
$or, b(b -30) -24(b -30) = 0$
$or, (b -24)(b -30) = 0$

Either,
$b -24 = 0$
$\therefore b = 24$

Or,
$b -30 = 0$
$\therefore b = 30$

Let b be the smaller side so, b = 24. Put value of b in equation (i), we get,
$or, l = 54 -24$
$\therefore l = 30$

So, l = 30m and b = 24m

In a square, the length is always equal to the breadth.
$or, l = b$
Let x be the part to be reduced from length of the rectangle.
$or, l - x = b$
$or, 30 - x = 24$
$or, x = 30-24$
$\therefore x = 6$

Now,

Percentage of the length to be decreased is given by:
$\dfrac{decreased \;part}{original length} × 100%$
$= \dfrac{x}{l} × 100%$
$= \frac{6}{30} × 100%$
$= 20%$

Therefore, the length of the rectangular field has to be decreased by 20% to make the given field a square.

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